Chapter 9.5, Problem 2GP

To determine

Find the solution.

Answer to Problem 2GP

  $x=6-\sqrt{41}$ or $x=6+\sqrt{41}$

Explanation of Solution

Given Information: $x^2-12x+3=8$

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

Calculation:

  $x^2-12x+3=8$

So,

  $x^2-12x+3-8=0$

Now, simplify

  $x^2-12x-5=0$

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

  $x^2-12x-5+{\left(\frac{12}{2}\right)}^2-{\left(\frac{12}{2}\right)}^2=0$

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

  ${\left(x-6\right)}^2-5-6^2=0$

So,

  ${\left(x-6\right)}^2-5-36=0$

So,

  ${\left(x-6\right)}^2-41=0$

Now, square root of 41 is $\sqrt{41}$

  ${\left(x-6\right)}^2-{\left(\sqrt{41}\right)}^2=0$

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

  $\left(x-6+\sqrt{41}\right)\left(x-6-\sqrt{41}\right)=0$

Hence,

  $x=6-\sqrt{41}$ or $x=6+\sqrt{41}$