Factoring Trinomials: Solving the equation

Let's solve the given equation:

$3x^2+9x-162=0$

$$First, note that the coefficients of all terms in the equation and the free term are divisible by 3 (this can also be determined for the free number - since the sum of its digits is divisible by 3), therefore we will simplify the equation first by dividing both sides by 3:

$3x^2+9x-162=0 \hspace{6pt}\text{/}:3\\ x^2+3x-54=0$

Now we notice that in the resulting equation the coefficient of the squared term is 1, therefore we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-54\\ m+n=3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the number 54 has several possible factor pairs, for example: 27 and 2, 9 and 6, etc. (we won't list them all here) but from the second requirement mentioned, along with the fact that the numbers we're looking for have different signs, leads to the conclusion that the difference between the absolute values of the pair of numbers we're looking for must be 3, therefore the only possibility for the two numbers we're looking for is:

$\begin{cases} m=9\\ n=-6 \end{cases}$

therefore we can factor the expression on the left side of the equation to:

$x^2+3x-54=0 \\ \downarrow\\ (x+9)(x-6)=0$

From here we'll remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown on one side:

$x+9=0\\ \boxed{x=-9}$

or:

$x-6=0\\ \boxed{x=6}$

Let's summarize the solution of the equation:

$3x^2+9x-162=0 \\ x^2+3x-54=0 \\ \downarrow\\ (x+9)(x-6)=0 \\ \downarrow\\ x+9=0\rightarrow\boxed{x=-9}\\ x-6=0\rightarrow\boxed{x=6}\\ \downarrow\\ \boxed{x=-9,6}$

Therefore the correct answer is answer D.

Let's solve the given equation:

$x(x^2-7x-8)=0$

Let's try to factor the expression inside of the parentheses:

$x^2-7x-8$

Note that in this expression the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-8\\ m+n=-7\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 8 are 2 and 4, or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=1 \end{cases}$

Therefore we'll factor the expression on the left side of the equation:

$x^2-7x-8\\ \downarrow\\ (x-8)(x+1)$

Now let's return to the original equation and apply the factoring that we completed earlier:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x+1)=0$

Remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain three simple equations and solve them by isolating the variable on one side:

$\boxed{x=0}$

or:

$x-8=0\\ \boxed{x=8}$

or:

$x+1=0\\ \boxed{x=-1}$

Let's summarize the solution of the equation:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x-3)=0 \\ \downarrow\\ \boxed{x=0}\\ x-8=0\rightarrow\boxed{x=8}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=0,8,-1}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$x^2-8x+15=3(x-3)$

$$First, let's organize the equation by opening the parentheses (using the extended distribution law) and combining like terms:

$x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-8x+15-3x+9=0 \\ x^2-11x+24=0$

Now we notice that in the resulting equation the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=24\\ m+n=-11\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 24 are 6 and 4, 12 and 2, 8 and 3, or 24 and 1. Meeting the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=-3 \end{cases}$

Therefore we can factor the expression on the left side of the equation to:

$x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain two simple equations and we'll proceed to solve them by isolating the variable in each of them:

$x-8=0\\ \boxed{x=8}$

or:

$x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=8,3}$

Therefore the correct answer is answer B.

Let's solve the following equation:

$-7(x+1)(x-3)(x+7)=0$

First, let's divide both sides of the equation by the number outside of the parentheses:

$-7(x+1)(x-3)(x+7)=0 \hspace{6pt}\text{/}:(-7)\\ (x+1)(x-3)(x+7)=0$

Remember that the product of an expression equals 0 only if at least one of the multiplying expressions equals zero,

Therefore we should obtain three simple equations and solve them by isolating the variable in each one:

$x+1=0\\ \boxed{x=-1}$or:

$x-3=0\\ \boxed{x=3}$or:

$x+7=0\\ \boxed{x=-7}$ Hence the solution to the equation is:

$\boxed{x=-1,3-7}$The correct answer is answer D.

To solve this problem, we'll follow these steps:

• Step 1: Eliminate the square roots by squaring both sides of the equation.
• Step 2: Simplify the resulting equation and bring all terms to one side.
• Step 3: Factor the resulting quadratic equation.
• Step 4: Solve for $x$ and check for valid solutions by substituting back into the original equation.

Now, let's work through each step in detail:

Step 1: Square both sides of the equation:

$\left(\sqrt{3x^2 + 10x - 16}\right)^2 = \left(\sqrt{2x^2 + 4x}\right)^2$

The equation becomes:

$3x^2 + 10x - 16 = 2x^2 + 4x$

Step 2: Simplify and rearrange the equation:

$3x^2 + 10x - 16 - 2x^2 - 4x = 0$

This simplifies to:

$x^2 + 6x - 16 = 0$

Step 3: Factor the quadratic equation:

We need to find two numbers that multiply to $-16$ and add to $6$. These numbers are $8$ and $-2$.

The equation factors to:

$(x + 8)(x - 2) = 0$

Step 4: Solve for $x$:

Set each factor equal to zero:

• $x + 8 = 0 \implies x = -8$
• $x - 2 = 0 \implies x = 2$

Finally, check these solutions in the original equation:

• For $x = -8$, $\sqrt{3(-8)^2 + 10(-8) - 16} = \sqrt{2(-8)^2 + 4(-8)}$

$\sqrt{192 - 80 - 16} = \sqrt{128 - 32}$

$\sqrt{96} = \sqrt{96}$

• For $x = 2$, $\sqrt{3(2)^2 + 10(2) - 16} = \sqrt{2(2)^2 + 4(2)}$

$\sqrt{12 + 20 - 16} = \sqrt{8 + 8}$

$\sqrt{16} = \sqrt{16}$

Both solutions are valid. Therefore, the solutions to the equation are:

$x = -8$ and $x = 2$.

The correct choice is $\textbf{d}: -8, 2$.

Let's solve the given equation:

$x+6=\frac{x^2+6}{x-4}$

We'll start by defining the domain of the unknown in the equation, this is because there is a variable in the denominator on the right side, we'll remember that the denominator cannot be zero and therefore we require that the expression in the denominator is not zero, meaning:

$x-4\neq0$

Next we'll solve the (point) inequality identically to how we solve a regular equation:

$x-4\neq0\\ \downarrow\\ \boxed{x\neq4}$

Therefore the domain of the unknown in the equation is:

$\textcolor{red}{\boxed{x\neq4}}$

(Note that any point inequality (and not slope inequality, meaning where $\neq$ and not \leq,\geq,>,< ) is solved identically to a regular equation)

Let's return to the given equation:

$x+6=\frac{x^2+6}{x-4}$

First we'll represent each term in the equation that isn't a fraction as a fraction, we'll do this using the fact that dividing any number by 1 doesn't change its value:

$x+6=\frac{x^2+6}{x-4} \\ \downarrow\\ \frac{ x}{1}+ \frac{6}{1}=\frac{x^2+6}{x-4}$

Next - we want to eliminate the fraction line, we'll do this by multiplying both sides of the equation by the simplest common denominator, in this case the simplest common denominator is the simple expression: $x-4$, this is because the other denominators are 1, therefore we'll multiply both sides of the equation by this expression, while remembering that the expressions in the numerator of each of the fractions in the equation will be multiplied by the expression that answers the question: "By how much did we multiply the current denominator to get the common denominator?" (for each fraction separately - see in the next calculation), then we'll open parentheses and simplify the equation by moving terms and combining like terms:

$\frac{x^{\diagdown\cdot(x-4)}}{1}+ \frac{6^{\diagdown\cdot(x-4)}}{1}=\frac{x^2+6^{\diagdown\cdot1}}{x-4} \text{/}\cdot(x-4) \\ \downarrow\\ x\cdot(x-4)+6\cdot(x-4)=(x^2+6)\cdot1\\ x^2-4x+6x-24=x^2+6 \\ 2x=30$

Let's continue, we got a first-degree equation, let's solve it in the regular way:

$2x=30 \hspace{6pt}\text{/}:2\\ \boxed{x=15}$

Now let's not forget the domain of the unknown in the equation:

$\boxed{x\neq4}$

Note that the solution we found does not contradict the domain and therefore the final solution is:

$\boxed{x=15}$

Let's summarize the solution of the equation:

$x+6=\frac{x^2+6}{x-4}\textcolor{red}{\rightarrow \boxed{x\neq4}} \\ \downarrow\\ \frac{ x}{1}+ \frac{6}{1}=\frac{x^2+6}{x-4}\\ \frac{x^{\diagdown\cdot(x-4)}}{1}+ \frac{6^{\diagdown\cdot(x-4)}}{1}=\frac{x^2+6^{\diagdown\cdot1}}{x-4} \text{/}\cdot(x-4) \\ \downarrow\\ x\cdot(x-4)+6\cdot(x-4)=(x^2+6)\cdot1\\ 2x=30 \rightarrow\boxed{x=15}\\ \downarrow\\ x=15\stackrel{?}{\textcolor{red}{\leftrightarrow} }\textcolor{red}{x\neq4}\\ \downarrow\\ \boxed{x\textcolor{green}{\stackrel{!}{=}}15}$

Therefore the correct answer is answer D.

Let's examine the problem:

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2=$

We'll focus on the expression under the square root. For calculation without a calculator, we naturally want to find a way to eliminate the square root. To do this, we need to transform the expression inside the root into a squared expression, which we'll try to do using the squared binomial formula in two main steps:

First - for simplicity of calculation and generality of solution let's denote:

$406=a\rightarrow405=a-1$

therefore the expression under the root is:

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+(a-1)^2a^2+a^2$

Step A.

We'll use a method called: "completing the square" In this method, we make the expression appear in the form of a squared binomial plus a correction term, meaning we "rearrange" the formula by moving terms within the formula itself, and use the standard form to get a structure similar to the squared binomial form:

$\boxed{ b^2-2bc+c^2= (b-c)^2}\\ \downarrow\\ \textcolor{blue}{b^2+c^2=(b-c)^2+2bc}$

Note that we can use this method on the expression in question (under the root), if we now denote:

$\textcolor{red}{ a-1=b\\ a=c }$

we get:

$(a-1)^2+(a-1)^2a^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ b^2+c^2+b^2c^2$

We'll use the "rearranged" formula we got earlier (highlighted in blue) and substitute it in the last expression we got:

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+a^2(a-1)^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ \textcolor{blue}{b^2+c^2}+b^2c^2 \\ \downarrow\\ \textcolor{blue}{(b-c)^2+2bc}+b^2c^2$

Now we'll return to parameter a using the definitions for b and c that we defined earlier (in red):

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+a^2(a-1)^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ \textcolor{blue}{b^2+c^2}+b^2c^2 \\ \downarrow\\ \textcolor{blue}{(b-c)^2+2bc}+b^2c^2 \\ \downarrow\\ ((a-1)-a)^2+2(a-1)a+(a-1)^2a^2$

where the first steps are just a reminder of what we've done so far and only the final step is substituting the expressions and returning to parameter a,

Let's continue and simplify the first term on the left in the expression we got:

$\textcolor{green}{ ((a-1)-a)^2}+2(a-1)a+(a-1)^2a^2 \\ \downarrow\\ \textcolor{green}{1}+2(a-1)a+(a-1)^2a^2$

Step B.

Now we'll use the laws of exponents:

$\boxed{x^ny^n=(xy)^2}$

and we'll express the third term from the left as a power:

$1+2(a-1)a+\textcolor{red}{(a-1)^2a^2} \\ 1+2(a-1)a+\textcolor{red}{((a-1)a)^2}$

Now let's rearrange, and factor using the shortened multiplication formula for squared binomial (in its addition form):

$1+2(a-1)a+((a-1)a)^2 \\ \downarrow\\ (\textcolor{orange}{(a-1)a})^2+2\textcolor{orange}{(a-1)a}\cdot\textcolor{purple}{1}+\textcolor{purple}{1}^2\\ \downarrow\\ (\textcolor{orange}{(a-1)a}+\textcolor{purple}{1})^2$

In the next step the square root will cancel out the squared power, which was actually the whole purpose of this development (let's summarize the solution steps):

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2= \\ \downarrow\\ \sqrt{(a-1)^2+(a-1)^2a^2+a^2 }-(a-1)^2= \\ \downarrow\\ \sqrt{\textcolor{red}{(a-1)^2+a^2}+(a-1)^2a^2 }-(a-1)^2= \\ \downarrow\\ \sqrt{\textcolor{red}{((a-1)-a)^2+2(a-1)a}+(a-1)^2a^2}-(a-1)^2\\ \downarrow\\ \sqrt{(\textcolor{orange}{(a-1)a})^2+2\textcolor{orange}{(a-1)a}\cdot\textcolor{purple}{1}+\textcolor{purple}{1}^2}-(a-1)^2\\ \downarrow\\ \sqrt{(\textcolor{orange}{(a-1)a}+\textcolor{purple}{1})^2}-(a-1)^2\\ \downarrow\\ \sqrt{((a-1)a+1)^2}-(a-1)^2= \\ \downarrow\\ (a-1)a+1-(a-1)^2$

We achieved our goal - we eliminated the root

Completing the solution:

Now we just need to expand the parentheses using the shortened multiplication formula for squared binomial and the distributive law, and then simplify the resulting expression:

$(a-1)a+1-(a-1)^2\\ \downarrow\\ a^2-a+1-(a^2-2a+1)\\ a^2-a+1-a^2+2a-1=\\ a$

Therefore we got that the result of simplifying the expression is simply:

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2= \\ \downarrow\\ \sqrt{(a-1)^2+a^2(a-1)^2+a^2 }-(a-1)^2= \\ a=406$

Therefore the correct answer is answer B

Note and additional question:

Note that the calculation was done in general form for parameter a, so now let's ask,

What will be the result of simplifying the expression?

$\sqrt{1414^2+1414^2\cdot1415^2+1415^2}-1414^2=?$