Chapter 9.5, Problem 47PFA

To determine

Find the solution.

Answer to Problem 47PFA

Option B as don’t need to subtract 25 from both sides

Explanation of Solution

Given Information: $x^2+10x+10=66$

Formula Used : $a^2+2ab+b^2={(a+b)}^2$ and $a^2-b^2=(a+b)(a-b)$

Calculation:

  $x^2+10x+10=66$

Now, subtract 10 from both sides.

  $x^2+10x+10-10=66-10$

Now, simplify

  $x^2+10x-56=0$

Now, adding & subtracting the square of half the coefficient of x.

  $x^2+10x-56+{\left(\frac{10}{2}\right)}^2-{\left(\frac{10}{2}\right)}^2=0$

Now use formula, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(x+5\right)}^2-56-{\left(5\right)}^2=0$

So,

  ${\left(x+5\right)}^2-56-25=0$

So,

  ${\left(x+5\right)}^2-81=0$

Now, square root of 81 is 9

  ${\left(x+5\right)}^2-{\left(9\right)}^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x+5+9\right)\left(x+5-9\right)=0$

Now,

  $\left(x+14\right)\left(x-4\right)=0$

Hence,

  $x=-14$ or $x=4$

Option B as don’t need to subtract 25 from both sides.