Solving Quadratic Equations by Completing the Square - Examples, Exercises and Solutions

 In order to calculate the value of the expression in the problem:

$a^2x^2-6ax+14$

Based on the given that:

$ax-3=1$

and without solving an equation (meaning - without finding the value of x, in this case),

we will use the method of completing the square,

First, let's recall the principles of this method and its general idea, and demonstrate it with a simpler expression:

In this method, we use the perfect square formulas in order to give the expression the form of a perfect square,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it the form of a perfect square,

That is, we use the formulas for perfect squares:

$(c\pm d)^2=c^2\pm2cd+d^2$

and we'll bring the expression to a square form by adding and subtracting the missing term,

For example:

Let's represent the expression:

$x^2-4x+3$

as a perfect square expression plus a correction,

First, we'll try to give the current expression a form that resembles the right side of the perfect square formulas mentioned, we'll also identify that we're interested in the subtraction form of the perfect square formula, since the non-squared term in the given expression has

a negative sign, we'll continue,

First, let's deal with the two terms with the highest powers in the expression:

$\underline{x^2-4x}+3 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{x^2-2\cdot x\cdot 2}+3 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2} \\$We identify that if we want to get from these two terms (underlined below in the calculation) a perfect square form,

We'll need to add to these two terms the term$2^2$, but we don't want to change the value of the expression in question, so we'll also subtract this number from the expression,

That is, we'll add and subtract the number (or expression) we need to "complete" to a perfect square form,

In the following calculation, the "trick" is demonstrated (two lines under the term we added and subtracted),

Later - we'll put it in perfect square form (demonstrated with colors) and in the final stage we'll further simplify the expression:

$x^2-2\cdot x\cdot 2+3\\ x^2-2\cdot x\cdot 2\underline{\underline{+2^2-2^2}}+3\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4+3\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{2})^2-4+3\\ \downarrow\\ \boxed{(x-2)^2-1}$

Therefore- we got the completing the square form for the given expression,

We won't expand here but we'll note that there are several different completion forms, since we can "rotate" the perfect square formulas in different ways,

Let's return then to the problem and examine again the expression on which we want to perform "completing the square":

$a^2x^2-6ax+14$We'll also remember that we want to use the information that:

$ax-3=1$Therefore, first we'll present the first term (which is the squared term in the expression) as a perfect square, and we'll examine the first two terms in comparison to the perfect square formula while we identify the expression form in the perfect square formula and try to understand what is the missing term in order to perform the completing the square:

$\underline{ a^2x^2-6ax} +14 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \underline{ (ax)^2-6ax} +14 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{ (\textcolor{green}{ax})^2\textcolor{red}{-2}\cdot \textcolor{green}{ax}\cdot\textcolor{purple}{3}} +14 \textcolor{blue}{\leftrightarrow} \underline{ c^2\textcolor{red}{-2}\textcolor{green}{c}\textcolor{purple}{d}\hspace{2pt}\boxed{+d^2} }\\$Again, we identify that in order to complete the expression of the first two terms to a square form, we need to add and subtract the number $3^2$ , we'll do this, then we'll get the square form using the perfect square formula and simplify the resulting expression:

$(ax)^2-2\cdot ax\cdot 3+14\\ (ax)^2-2\cdot ax\cdot 3\underline{\underline{+3^2-3^2}}+14\\ (\textcolor{red}{ax})^2-2\cdot \textcolor{red}{ax}\cdot \textcolor{green}{2}+\textcolor{green}{3}^2-9+14\\ \downarrow\\ (\textcolor{red}{ax}-\textcolor{green}{3})^2-9+14\\ \downarrow\\ \boxed{(ax-3)^2+5}$

To summarize, we got therefore that:

$a^2x^2-6ax +14=\\ (ax)^2-2\cdot ax\cdot3+14=\\ (ax-3)^2-9+14=\\ \boxed{(ax-3)^2}+5$

Now we'll use the given that:

$ax-3=1$

We'll use this given "completely" and substitute its value in the expression we got:

$a^2x^2-6ax +14=\\ (\underline{ax-3})^2+5 =\text{?}\leftrightarrow \underline{ax-3}=1\\ \downarrow \\ \underline{1}^2+5=\\ \boxed{6}$

Therefore the correct answer is answer C.

 First, let's recall the principles of the "completing the square" method and its general concept:

In this method, we use the perfect square formulas in order to give an expression the form of a perfect square,

This method is called "completing the square" because in this method we "complete" a missing part of a certain expression in order to get from it a perfect square form,

That is, we use the perfect square formulas:

$(c\pm d)^2=c^2\pm2cd+d^2$

and we bring the expression to a perfect square form by adding and subtracting the missing term,

In the given problem we will first look at the given equation:

$16x^2+24x-40=0$

First, we'll try to give the expression on the left side of the equation a form that resembles the right side of the perfect square formulas mentioned above, we also identify that we are interested in the addition form of the perfect square formula, since the non-square term in the given expression, $24x$has a positive sign,

Let's continue,

First, let's deal with the two terms with the highest powers in the expression on the left side of the equation,

and we'll try to identify the missing term by comparing it to the perfect square formula,

To do this - first we'll present these terms in a form similar to the form of the first two terms in the perfect square formula:

$\underline{16x^2+24x}-40 \textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{(\textcolor{red}{4x})^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{4x}\cdot \textcolor{green}{3}}-40 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the perfect square formula (on the right side of the blue press in the previous calculation) we are actually making the analogy:

$\begin{cases} 4x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we can identify that if we want to get a perfect square form from these two terms (underlined below in the calculation),

We will need to add to these two terms the term$3^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

That is, we'll add and subtract the term (or expression) we need to "complete" to a perfect square form,

In the next calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next, we'll put into perfect square form the appropriate expression (demonstrated with colors) and in the final stage we'll further simplify the expression:

$(4x)^2+2\cdot 4x\cdot 3-40\\ (4x)^2+2\cdot 4x\cdot 3\underline{\underline{+3^2-3^2}}-40\\ (\textcolor{red}{4x})^2+2\cdot \textcolor{red}{4x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9-40\\ \downarrow\\ (\textcolor{red}{4x}+\textcolor{green}{3})^2-9-40\\ \downarrow\\ \boxed{(4x+3)^2-49}$

Therefore- we got the completing the square form for the given expression,

Let's summarize the development stages, we'll do this now within the given equation:

$16x^2+24x-40=0\\ (4x)^2+2\cdot 4x\cdot 3\underline{\underline{+3^2-3^2}}-40=0\\ (\textcolor{red}{4x})^2+2\cdot \textcolor{red}{4x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9-40=0\\ \downarrow\\ (\textcolor{red}{4x}+\textcolor{green}{3})^2-9-40=0\\ \downarrow\\ \boxed{(4x+3)^2-49=0}$

Let's note now that we are interested in the value of the expression:

$12x+9=\text{?}$

But how is this expression related to the expression we got in the last stage?

We can notice the connection if we also refer to the fact that the expression whose value we want to calculate can be factored by taking out a common factor to get the expression:

$$$12x+9=\text{?} \\ \downarrow\\ 3(4x+3)=\text{?}$

That is, to answer the question asked, it's enough for us to find the value of the expression:

$4x+3=\text{?}$

But the value of this expression we can easily get from the equation we got in the last stage (that is, after performing the "completing the square"), we'll do this by isolating the squared expression on one side and then taking the square root:

$(4x+3)^2-49=0\\ (4x+3)^2=49\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ 4x+3=\pm7$

(We'll remember of course that taking the square root from both sides of the equation involves considering two possibilities - with a positive and negative sign)

Therefore, now we can complete calculating the value of the requested expression, we'll do this by substituting the value of the expression we just calculated (highlighted in color in the next calculation) in the expression whose value we want to calculate:

$12x+9= \\ 3(\textcolor{orange}{4x+3})=\\ 3\cdot(\textcolor{orange}{\pm7})=\\ \pm21 \\ \downarrow\\ \boxed{12x+9\stackrel{\textcolor{red}{!}}{=}\pm21}$

Therefore the correct answer is answer A.

Let's first recall the principles of the "completing the square" method and its general idea:

In this method, we use the quadratic binomial formulas in order to give an expression the form of a quadratic binomial,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it a form of a quadratic binomial,

That is, we use the formulas for quadratic binomial:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we bring the expression to a quadratic form by adding and subtracting the missing term,

In the given problem let's first look at the given equation:

$25x^2+30x+6=0$

First, we'll try to give the expression on the left side of the equation a form that resembles the right side of the quadratic binomial formulas, we also identify that we are interested in the addition form of the quadratic binomial formula, since the non-squared term in the given expression, 30x, has a positive sign.

Let's continue,

First, let's deal with the two terms with the highest powers in the expression on the left side of the equation,

And we'll try to identify the missing term by comparing it to the quadratic binomial formula,

For this - first we'll present these terms in a form similar to the form of the first two terms in the quadratic binomial formula:

$\underline{ 25x^2+30x}+6\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{(\textcolor{red}{5x})^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{5x}\cdot \textcolor{green}{3}}+6 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

We can notice that in comparison to the quadratic binomial formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} 5x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a quadratic binomial form,

We will need to add to these two terms the term

$3^2$

However, we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

In other words, we'll add and subtract the term (or expression) we need to "complete" to a quadratic binomial form,

The following calculation demonstrates the "trick" (two lines under the term we added and subtracted from the expression),

Next - we'll put into quadratic binomial form the appropriate expression (demonstrated with colors) and in the final stage we'll further simplify the expression:

$(5x)^2+2\cdot 5x\cdot 3+6\\ (5x)^2+2\cdot5x\cdot 3\underline{\underline{+3^2-3^2}}+6\\ (\textcolor{red}{5x})^2+2\cdot \textcolor{red}{5x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9+6\\ \downarrow\\ (\textcolor{red}{5x}+\textcolor{green}{3})^2-9+6\\ \downarrow\\ \boxed{(5x+3)^2-3}$

Therefore- we got the completing the square form for the given expression,

Let's summarize the development stages, we'll do this now within the given equation:

$25x^2+30x+6=0 \\ (5x)^2+2\cdot 5x\cdot 3+6=0\\ (\textcolor{red}{5x})^2+2\cdot \textcolor{red}{5x}\cdot \textcolor{green}{3}\underline{\underline{+\textcolor{green}{3}^2-3^2}}+6=0\\ \downarrow\\ (\textcolor{red}{5x}+\textcolor{green}{3})^2-9+6=0\\ \downarrow\\ \boxed{(5x+3)^2-3=0}$

Let's notice now that we are interested in the value of the expression:

$5x+3=\text{?}$

Therefore, we can return to the equation we got in the last stage and isolate this expression from it,

We'll do this by moving terms and taking the square root:

$(5x+3)^2-3=0\\ (5x+3)^2=3\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{5x+3=\pm\sqrt{3}}$

(We'll remember of course that taking the square root from both sides of an equation involves considering two possibilities - with a positive and negative sign)

Therefore the correct answer is answer D.

First, let's recall the principles of the "completing the square" method and its general idea:

In this method, we use the formulas for the square of a binomial in order to give an expression the form of a squared binomial,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it a form of a squared binomial,

That is, we use the formulas for the square of a binomial:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we bring the expression to a squared form by adding and subtracting the missing term,

In the given problem we will first refer to the given equation:

$121x^2-44x-9=0$

First, we will try to give the expression on the left side of the equation a form that resembles the form of the right side in the abbreviated multiplication formulas mentioned, we will also identify that we are interested in the subtraction form of the abbreviated multiplication formula, this is because the non-squared term in the given expression, 44x is negative, we will continue,

First, we will deal with the two terms with the highest powers in the expression requested which is on the left side of the equation,

And we will try to identify the missing term in comparison to the abbreviated multiplication formula,

To do this- first we will present these terms in a form similar to the form of the first two terms in the abbreviated multiplication formula:

$\underline{ 121x^2-44x}-9\textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \\ \hspace{4pt}\\ \\ \downarrow\\ \underline{(\textcolor{red}{11x})^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}}-9 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

It can be noticed that in comparison to the abbreviated multiplication formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} 11x\textcolor{blue}{\leftrightarrow}c\\ 2\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we will identify that if we want to get a squared binomial form from these two terms (underlined below in the calculation),

We will need to add to these two terms the term

$2^2$

However, we don't want to change the value of the expression in question, and therefore- we will also subtract this term from the expression,

That is, we will add and subtract the term (or expression) we need to "complete" to the form of a squared binomial,

In the following calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next- we will put into the squared binomial form the appropriate expression (demonstrated with colors) and in the last stage we will further simplify the expression:

$(11x)^2-2\cdot 11x\cdot 2-9\\ (11x)^2-2\cdot11x\cdot 2\underline{\underline{+2^2-2^2}}-9\\ (\textcolor{red}{11x})^2-2\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4-9\\ \downarrow\\ (\textcolor{red}{11x}-\textcolor{green}{2})^2-4-9\\ \downarrow\\ \boxed{(11x-2)^2-13}$

Hence- we obtained the completing the square form for the given expression,

Let's summarize the development stages, we will do this now within the given equation:

$121x^2-44\sqrt{2}x-9=0 \\ (11x)^2-2\cdot 11x\cdot 2-9=0\\ (\textcolor{red}{11x})^2-2\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}\underline{\underline{+\textcolor{green}{2}^2-2^2}}-9=0\\ \downarrow\\ (\textcolor{red}{11x}-\textcolor{green}{2})^2-4-9=0\\ \downarrow\\ \boxed{(11x-2)^2-13=0}$

Now, we can isolate from this expression a simpler algebraic expression,

We will do this by moving terms and extracting a square root:

$(11x-2)^2-13=0\\ (11x-2)^2=13\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{11x-2=\pm\sqrt{13}}$

(We should remember of course that extracting a square root from both sides of the equation involves considering two possibilities - with a positive sign and with a negative sign)

Let's note now that we are interested in the value of the expression:

$11x+9=\text{?}$

Which we will easily extract from the equations that we obtained,

At this stage we will emphasize two important things:

A. We obtained two equations requiring two values with opposite signs for the same expression:

$11x-2=\pm\sqrt{13}$

However it's easy to understand that these two equations cannot be held together unless the expression equals 0, which is not the case here.

B. Due to this fact, we need to separate and solve individually in order to obtain all the possibilities for the value of the requested expression,

We will continue, and refer to each equation separately, first we will try to identify the requested expression, and then isolate it, in each equation separately:

$11x-2=\pm\sqrt{13} \\ \underline{\textcolor{blue}{11x+9}}-11=\pm\sqrt{13} \\ \downarrow\\ 11x+9-11=\sqrt{13} \rightarrow\boxed{11x+9=11+\sqrt{13}} \\ 11x+9-11=-\sqrt{13}\rightarrow\boxed{11x+9=11-\sqrt{13}} \\ \downarrow\\ \boxed{11x+9=11+\sqrt{13},\hspace{4pt}11-\sqrt{13}}$

Therefore, the correct answer is answer A.

We want to calculate the value of the expression:

$x^2+\frac{1}{x^2}=\text{?}$

based on the given equation:

$x+\frac{1}{x}=4$

but without solving it for x,

For this, let's first note that while the given equation deals with terms with first power only,

in the expression whose value we want to calculate - there are terms with second power only,

therefore we understand that apparently we need to square the expression on the left side of the given equation,

We'll remember of course the shortened multiplication formula for a binomial square:

$(a\pm b)^2=a^2\pm2ab+b^2$

and we'll square both sides of the given equation, later we'll emphasize something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its inverse are added):

$x+\frac{1}{x}=4 \hspace{6pt}\text{/}()^2\\ (x+\frac{1}{x})^2=4^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{1}{x}}+ \frac{1}{x^2}=16\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{1}+ \frac{1}{x^2}=16\\$Let's now notice that the "mixed" term in the shortened multiplication formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning - it's not dependent on the variable x, since it's a multiplication between an expression and its inverse,

This fact actually allows us to isolate the desired expression from the equation we get and find its value (which is not dependent on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2+2\cdot \textcolor{blue}{1}+ \frac{1}{x^2}=16\\ x^2+2+ \frac{1}{x^2}=16\\ \boxed{x^2+\frac{1}{x^2}=14}$

Therefore the correct answer is answer D.

Our goal is to calculate the value of the following expression:

$x^2+\frac{9}{x^2}=\text{?}$

based on the given equation:

$x+\frac{3}{x}=5$

However without solving it for x,

Note that while the given equation deals with terms raised to the first power only,

in the expression we want to calculate - there are terms raised to the second power.

Therefore we need to square the expression on the left side of the given equation.

We can do this by using the shortened multiplication formula for binomial square:

$(a\pm b)^2=a^2\pm2ab+b^2$

We'll proceed to square both sides of the given equation, later on we'll discuss what happens in the given mathematical structure when a term and its proportional inverse are added:

$x+\frac{3}{x}=5 \hspace{6pt}\text{/}()^2\\ (x+\frac{3}{x})^2=5^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{3}{x}}+ \frac{3^2}{x^2}=25\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{3}+ \frac{9}{x^2}=25\\$For now take note that the "mixed" term in the shortened multiplication formula ($2ab$) gives us - as a result of squaring the mathematical structure in question - a free number. This signifies that it's not dependent on the variable x, since it involves multiplication between an expression with a variable and its proportional inverse.

This fact actually allows us to isolate the desired expression from the equation we obtain. We can subsequently determine its value (which is not dependent on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2+2\cdot \textcolor{blue}{3}+ \frac{9}{x^2}=25\\ x^2+6+ \frac{9}{x^2}=25\\ \boxed{x^2+\frac{9}{x^2}=19}$

Therefore the correct answer is answer C.

First, let's recall the principles of the "completing the square" method and its general idea:

In this method, we use the formulas for the square of a binomial in order to give an expression the form of a squared binomial,

This method is called "completing the square" due to the fact that in this method we "complete" a missing part of a certain expression in order to obtain from it a form of a squared binomial,

That is, we use the formulas for the square of a binomial:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we bring the expression to a squared form by adding and subtracting the missing term,

In the given problem we will first refer to the given equation

$2x^2+14\sqrt{2}x-15=0$

First, we will try to give the expression on the left side of the equation a form that resembles the form of the right side in the abbreviated multiplication formulas mentioned, we will also identify that we are interested in the addition form of the abbreviated multiplication formula, this is because the term that is not squared in the given expression,:

$14\sqrt{2}x$

has a positive sign,we will continue,

First, we will deal with the two terms with the highest powers in the expression requested on the left side of the equation,

And we will try to identify the missing term in comparison to the abbreviated multiplication formula,

To do this- first we will present these terms in a form similar to the form of the first two terms in the abbreviated multiplication formula:

$\underline{ 2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \hspace{4pt}\\ \\ \underline{ (\sqrt{2})^2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \downarrow\\ \underline{(\textcolor{red}{\sqrt{2}x})^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}}-15 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

It can be noticed that in comparison to the abbreviated multiplication formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} \sqrt{2}x\textcolor{blue}{\leftrightarrow}c\\ 7\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get a squared binomial form from these two terms (underlined below in the calculation),

we will need to add to these two terms the term $7^2$

However, we don't want to change the value of the expression in question, and therefore- we will also subtract this term from the expression,

That is, we will add and subtract the term (or expression) we need to "complete" to the form of a squared binomial,

In the next calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Then- we will put into the squared binomial form the appropriate expression (demonstrated with colors) and in the last stage we will further simplify the expression:

$(\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15\\ (\sqrt{2}x)^2+2\cdot\sqrt{2}x\cdot 7\underline{\underline{+7^2-7^2}}-15\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}+\textcolor{green}{7}^2-49-15\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64}$

Thus- we obtained the completing the square form for the given expression,

Let's summarize the development stages, we will do this now within the given equation:

$2x^2+14\sqrt{2}x-15=0 \\ (\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15=0\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}\underline{\underline{+\textcolor{green}{7}^2-7^2}}-15=0\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15=0\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64=0}$

Now, we can isolate from this expression a simpler algebraic expression,

We will do this by transferring sides and extracting a square root:

$(\sqrt{2}x+7)^2-64=0\\ (\sqrt{2}x+7)^2=64\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{\sqrt{2}x+7=\pm8}$

(We will remember of course that extracting a square root from both sides of the equation involves considering two possibilities - with a positive sign and with a negative sign)

Let's note now that we are interested in the value of the expression:

$\sqrt {2}x+5=\text{?}$

Which we can easily extract from the equations we obtained,

At this stage we will emphasize two important things:

A. We obtained two equations requiring two values with opposite signs for the same expression

$\sqrt{2}x+7=\pm8$

But it's easy to understand that these two equations cannot be held together unless the expression equals 0, which is not the case here.

B. Because of this, we need to separate and solve each one independently in order to get all the possibilities for the value of the requested expression,

We will continue, and refer to each equation separately, first we will try to identify the requested expression, and then isolate it, in each equation separately:

$\sqrt{2}x+7=\pm8 \\ \underline{\textcolor{blue}{\sqrt{2}x+5}}+2=\pm8 \\ \downarrow\\ \sqrt{2}x+5+2=8 \rightarrow\boxed{\sqrt{2}x+5=6} \\ \sqrt{2}x+5+2=-8 \rightarrow\boxed{\sqrt{2}x+5=-10} \\ \downarrow\\ \boxed{\sqrt{2}x+5=6,\hspace{4pt}-10}$

Therefore, the correct answer is answer D.

Let's recall the shift rule which states that for the parabola:

$y=a(x-x_1)^2+y_1$

We get from the parabola:

$y=ax^2$

By shifting it (meaning - shifting all points on it)

  x_1

$y_1$ units (positive) to the right, and units (positive) upward,

(We'll also remember that for negative x1 or y1 - then the shift (in positive units) is in the opposite direction according to this rule),

Therefore, we want to express the parabola, after the shift:

$y=x^2-6x-5$

In the form:

$y=(x-x_1)^2+y_1$

For this we'll use completing the square -

Let's first recall the principles of the "completing the square" method and its general concept:

In this method we use the shortened multiplication formulas for binomial squares in order to give an expression the form of a binomial square,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it a binomial square form,

Meaning we use the shortened formulas for binomial squares:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we'll bring the expression to a square form by adding and subtracting the missing term,

In the given problem we'll first look at the given parabola after the shift:

$y=x^2-6x-5$

First, we'll try to give the parabola expression a form that resembles the right side of the shortened multiplication formulas, we'll also identify that we're interested in the subtraction form of the shortened multiplication formula, since the non-squared term in the given expression, -6X, has a negative sign,we'll continue, first dealing with the two terms with the highest powers in the expression which are on the left side of the equation,

And we'll try to identify the missing term by comparing to the shortened multiplication formula,

For this - first we'll present these terms in a form similar to the form of the first two terms in the shortened multiplication formula:

$y= \underline{x^2-6x}-5 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ y=\underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}}-5 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

We can notice that compared to the shortened multiplication formula (right side of the blue press in the previous calculation) we're actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term$3^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning- we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

The following calculation demonstrates the "trick" (two lines under the term we added and subtracted from the expression),

Next - we'll put into binomial square form the appropriate expression (demonstrated using colors) and in the final stage we'll further simplify the expression:

$y= x^2-2\cdot x\cdot 3-5\\ (3x)^2-2\cdot x\cdot 3\underline{\underline{+3^2-3^2}}-5\\ y=\textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9-5\\ \downarrow\\ y= (\textcolor{red}{x}-\textcolor{green}3)^2-9-5\\ \downarrow\\ \boxed{y=(x-3)^2-14}$

Therefore- we got the completing the square form for the expression in the parabola after the shift,

We can now compare to the general form in the shift rule that we mentioned earlier:

$y=(x-x_1)^2+y_1 \\ \updownarrow\\ y=(x-3)^2-14$

We identify therefore using the shift rule we mentioned at the beginning of the solution, that we got this parabola from shifting the parabola:

$y=x^2$

3 units right and 14 units down, therefore the correct answer is answer C

 Let's first recall the principles of the "completing the square" method and its general idea:

In this method, we use the perfect square formulas in order to give an expression the form of a perfect square,

This method is called "completing the square" because in this method we "complete" a missing part of a certain expression in order to get from it a perfect square form,

That is, we use the formulas for perfect squares:

$(c\pm d)^2=c^2\pm2cd+d^2$

and we bring the expression to a square form by adding and subtracting the missing term,

In the given problem let's first look at the given equation:

$9x^2-30x+4=0$

First, we'll try to give the expression on the left side of the equation a form that resembles the right side of the perfect square formulas mentioned above, we also identify that we are interested in the subtraction form of the perfect square formula, because the non-square term in the given expression, $$$30x$,

has a negative sign,

Let's continue,

First, let's deal with the two terms with the highest powers in the expression on the left side of the equation,

and we'll try to identify the missing term by comparing it to the perfect square formula,

To do this- first we'll present these terms in a form similar to the form of the first two terms in the perfect square formula:

$\underline{9x^2-30x}+4 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{(\textcolor{red}{3x})^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{3x}\cdot \textcolor{green}{5}}+4 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the perfect square formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} 3x\textcolor{blue}{\leftrightarrow}c\\ 5\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get a perfect square form from these two terms (underlined in the calculation),

We will need to add to these two terms the term$5^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

In other words- we'll add and subtract the term (or expression) we need to "complete" to a perfect square form,

The next calculation demonstrates the "trick" (two lines under the term we added and subtracted from the expression),

Next- we'll put into perfect square form the appropriate expression (demonstrated using colors) and in the final stage we'll simplify the expression further:

$(3x)^2-2\cdot 3x\cdot 5+4\\ (3x)^2-2\cdot 3x\cdot 5\underline{\underline{+5^2-5^2}}+4\\ (\textcolor{red}{3x})^2-2\cdot \textcolor{red}{3x}\cdot \textcolor{green}{5}+\textcolor{green}{5}^2-25+4\\ \downarrow\\ (\textcolor{red}{3x}-\textcolor{green}{5})^2-25+4\\ \downarrow\\ \boxed{(3x-5)^2-21}$

Therefore- we got the completing the square form for the given expression,

Let's summarize the development steps, we'll do this now within the given equation:

$9x^2-30x+4=0 \\ (3x)^2-2\cdot 3x\cdot 5+4=0\\ (\textcolor{red}{3x})^2-2\cdot \textcolor{red}{3x}\cdot \textcolor{green}{5}\underline{\underline{+\textcolor{green}{5}^2-5^2}}+4=0\\ \downarrow\\ (\textcolor{red}{3x}-\textcolor{green}{5})^2-25+4=0\\ \downarrow\\ \boxed{(3x-5)^2-21=0}$

Let's note now that we are interested in the value of the expression:

$3x-5=\text{?}$

Therefore, we can return to the equation we got in the last stage and isolate this expression from it,

We'll do this by moving terms and taking the square root:

$(3x-5)^2-21=0\\ (3x-5)^2=21\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{3x-5=\pm\sqrt{21}}$

(We'll remember of course that taking the square root from both sides of an equation involves considering two possibilities - with a positive and negative sign)

Therefore the correct answer is answer C.

We want to calculate the value of c in the equation:

$16x^2+\frac{9}{x^2}=c$

based on the given equation:

$x-\frac{3}{4x}=1$

but without solving it for x,

For this, let's first notice that while the given equation deals with terms to the first power only,

in the expression we want to calculate:

$16x^2+\frac{9}{x^2}=\text{?}$

there are terms to the second power only,

therefore we understand that apparently we need to square the left side of the given equation,

We'll remember of course the formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

and we'll square both sides of the given equation, later we'll highlight something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its reciprocal are added):

$x-\frac{3}{4x}=1 \hspace{6pt}\text{/}()^2\\ (x-\frac{3}{4x})^2=1^2\\ \downarrow\\ x^2-2\cdot \textcolor{blue}{x\cdot\frac{3}{4x}}+ \frac{3^2}{(4x)^2}=1\\ \downarrow\\ x^2-2\cdot \textcolor{blue}{\frac{3}{4}}+ \frac{9}{16x^2}=1\\$Let's now notice that the "mixed" term in the binomial square formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning- one that doesn't depend on the variable x, since it involves multiplication between an expression with a variable and its reciprocal,

This fact actually allows us to isolate the desired expression (or close to it) from the equation we get and find its value (which doesn't depend on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2-2\cdot \textcolor{blue}{\frac{3}{4}}+ \frac{9}{16x^2}=1\\ x^2-\frac{6}{4}+ \frac{9}{16x^2}=1\\ x^2-\frac{3}{2}+ \frac{9}{16x^2}=1\\ \boxed{x^2+ \frac{9}{16x^2}=\frac{5}{2}}$

Let's continue and focus on our goal, finding c in the equation:

$16x^2+\frac{9}{x^2}=c$

Note that in order to get the left side in the equation mentioned from the left side in the equation we reached earlier, we'll need to multiply both sides of the equation by 16:

$x^2+ \frac{9}{16x^2}=\frac{5}{2}\hspace{6pt}\text{/}\cdot16\\ 16x^2+\frac{\not{16}\cdot9}{\not{16}x^2}=16\cdot\frac{5}{2}\\ 16x^2+\frac{9}{x^2}=40\\ \updownarrow\\ 16x^2+\frac{\cdot9}{x^2}=c\leftrightarrow c=\text{?}\\ \downarrow\\ \boxed{c=40}$

Now, let's try to answer the additional question asked:

Is every solution of the second equation (with c) also a solution to the given equation?

In other words-

Is each one of the solutions to the equation:

$16x^2+\frac{\cdot9}{x^2}=40$

also a solution to the given equation:

$x-\frac{3}{4x}=1$?

Let's note that we reached the first equation mentioned here from the second equation mentioned (which is the given equation) by squaring both sides of the given equation and moving terms,

Generally- equations that are derived from one another through moving terms, multiplying or dividing by a constant are equivalent, meaning their solutions are identical,

However- equations derived from one another by raising to an even power are not necessarily equivalent, because raising to a square (or any even power), might, since it always yields a non-negative result, add solutions to the equation.

Therefore, we cannot determine (without solving the equation for the unknown) whether every solution to the equation:

$16x^2+\frac{\cdot9}{x^2}=40$

is necessarily also a solution to the equation:

$x-\frac{3}{4x}=1$

and therefore the correct answer is answer B.

We want to calculate the value of the expression:

$x^2+\frac{49}{x^2}=\text{?}$

based on the given equation:

$x+\frac{7}{x}=2$

but without solving it for x,

For this, let's first note that while the given equation deals with terms in first power only,

in the expression we want to calculate - there are terms in second power only,

therefore we understand that apparently we need to square the expression on the left side of the given equation,

We'll remember of course the formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

and we'll square both sides of the given equation, later we'll emphasize something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its proportional inverse are added):

$x+\frac{7}{x}=2 \hspace{6pt}\text{/}()^2\\ (x+\frac{7}{x})^2=2^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{7}{x}}+ \frac{7^2}{x^2}=4\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{7}+ \frac{49}{x^2}=4\\$Let's now notice that the "mixed" term in the square formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning - it's not dependent on variable x, since it involves multiplication between an expression with a variable and its proportional inverse,

This fact actually allows us to isolate the desired expression from the equation we got and get its value (which is not dependent on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2+2\cdot \textcolor{blue}{7}+ \frac{49}{x^2}=4\\ x^2+14+ \frac{49}{x^2}=4\\ \boxed{x^2+\frac{49}{x^2}=-10}$

Now, let's try to answer the additional question asked:

How many (real) solutions does the given equation have?,

For this, let's examine the equation we got by squaring the given equation and by moving terms between sides,

Let's note that on the left side there's an expression that is a sum of two positive terms, and therefore it is certainly an expression that has only positive values:

x^2+\frac{49}{x^2}\rightarrow x^2,\hspace{4pt}\frac{49}{x^2}\\ \downarrow\\ x^2>0 \hspace{6pt}(x\neq0)\\ \frac{49}{x^2}>0 \hspace{6pt}(x\neq0)\\ \downarrow\\ \boxed{x^2+\frac{49}{x^2}>0}

This is because even powers will always give positive results or zero (which in this case is ruled out by the domain of definition of the unknown in the equation), and since the sum of two positive terms is positive too,

We'll continue and examine the right side of the resulting equation, and conclude that this is impossible, since the resulting equation requires that an expression that is certainly positive (on the left side), to be negative (the expression on the right side):

\textcolor{red}{x^2+\frac{49}{x^2}}>0\leftrightarrow\textcolor{red}{x^2+\frac{49}{x^2}=-10}\\ \updownarrow(\text{but})\\\ \textcolor{red}{-10}<0 $$

Therefore there is no (real) value of the unknown x that when substituted in the equation:

$x^2+\frac{49}{x^2}=-10$will give a true statement.

And certainly any solution to the given equation must satisfy the equation we got by squaring both sides (mentioned above),

Therefore- there is no (real) solution to the given equation.

(And we concluded this without trying to solve it for the unknown x)

Therefore the correct answer is answer C.

Let's recall first that the equation of a circle with center at point$O(x_o,y_o)$ and radius $R$is:

$(x-x_o)^2+(y-y_o)^2=R^2$

Let's now return to the problem and the given circle equation and examine it:

$x^2-8x+y^2-6y=24$

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that the right side contains the sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

For this, first let's recall the binomial square formulas:

$(c\pm d)^2=c^2\pm2cd+d^2$

and let's deal separately with the part of the equation related to x (underlined):

$\underline{ x^2-8x}+y^2-6y=24$

Let's continue, for convenience and clarity of discussion - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the binomial square formula (we'll choose the subtraction form of the binomial square formula since the first-degree term in the expression we're dealing with$8x$ has a negative sign):

$\underline{ x^2-8x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the binomial square formula (from the right side of the blue press in the previous calculation) we are actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we can identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term$4^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning, we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

In the following calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next, we'll put into binomial square form the appropriate expression (demonstrated with colors) and in the final stage we'll further simplify the expression:

$x^2-2\cdot x\cdot 4\\ x^2-2\cdot x\cdot 4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4})^2-16}\\$Let's summarize the development stages so far for the x-related expression, we'll do this now within the given equation:

$x^2-8x+y^2-6y=24 \\ x^2-2\cdot x\cdot 4+y^2-6y=24 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-6y=24\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4})^2-16+y^2-6y=24\\$We'll continue and perform an identical process for the y-related terms in the resulting equation:

$(x-4)^2-16+\underline{y^2-6y}=24\\ \downarrow\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3}=24\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3\underline{\underline{+3^2-3^2}}}=24\\ \downarrow\\ (x-4)^2-16+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=24\\ \downarrow\\ (x-4)^2-16+(\textcolor{red}{y}-\textcolor{green}{3})^2-9=24\\ \boxed{(x-4)^2+(y-3)^2=49}$