Solving Quadratic Equations by Completing the Square: Finding the value of a quadratic expression based on an equation

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.

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.

 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 the expression:

$x^3+\frac{1}{x^3}=\text{?}$

Based on the given equation:

$x+\frac{1}{x}=5$

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 we want to calculate - there are terms with third power only,

However to start let's note additionally that the value of the expression:

$x^2+\frac{1}{x^2}=\text{?}$can be calculated more easily, since it involves only terms with second power:

Therefore, we understand that to do this - apparently we need to square the expression on the left side of the given equation,

We'll recall 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, then 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 opposite are added):

$x+\frac{1}{x}=5 \hspace{6pt}\text{/}()^2\\ (x+\frac{1}{x})^2=5^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{1}{x}}+ \frac{1}{x^2}=25\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{1}+ \frac{1}{x^2}=25\\$Let's now note that the "mixed" term in the 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's a multiplication between an expression and its opposite,

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

$x^2+2\cdot \textcolor{blue}{1}+ \frac{1}{x^2}=25\\ x^2+2+ \frac{1}{x^2}=25\\ \boxed{x^2+\frac{1}{x^2}=23}$

From here we'll continue and return to our goal- calculating the value of the expression:

$x^3+\frac{1}{x^3}=\text{?}$Let's note that if we multiply both sides of the last equation we got by the expression: $x+\frac{1}{x}$we can get on the left side an expression containing the desired terms (with third power) and additionally the value of this expression we already know since the given equation is:

$\textcolor{blue}{x+\frac{1}{x}=5}$Let's do this then, and afterwards we'll try to isolate the desired expression from the resulting equation:

$$$x^2+\frac{1}{x^2}=23\hspace{6pt}\text{/}\cdot(x+\frac{1}{x})\\ (x+\frac{1}{x})(x^2+\frac{1}{x^2})=(x+\frac{1}{x})\cdot23$

From here we'll continue in two different ways, on the left side, from which we want to get the desired terms (with third power), we'll prefer to expand the parentheses using the expanded distribution law and simplify later, but on the right side, where we don't want to have dependency on the unknown, we'll substitute what we know from the given equation (highlighted in blue):

$(x+\frac{1}{x})(x^2+\frac{1}{x^2})=\textcolor{blue}{(x+\frac{1}{x})}\cdot23\\ \downarrow\\ x^3+\frac{1}{x}+x+\frac{1}{x^3}=\textcolor{blue}{5}\cdot23\\ x^3+\frac{1}{x^3}+x+\frac{1}{x}=115\\$$$Now we'll identify again that the sum of the last two terms in the expression on the left side - we already know (again - from the given equation - highlighted in blue), and therefore we can substitute this information again, isolate the desired expression and get its value - without dependency on the unknown x:

$x^3+\frac{1}{x^3}+\textcolor{blue}{\underline{x+\frac{1}{x}}}=115\\ \downarrow\\ x^3+\frac{1}{x^3}+\textcolor{blue}{\underline{5}}=115\\ \boxed{x^3+\frac{1}{x^3}=110}$

Therefore the correct answer is answer B.