Chapter 9.5, Problem 23PPS

To determine

Find the solution.

Answer to Problem 23PPS

$x=-0.2$ or $x=11.2$

Explanation of Solution

Given Information: $x^2-11x+3=5$

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

Calculation:

  $x^2-11x+3=5$

So, simplify

  $x^2-11x+3-5=0$

Now,

  $x^2-11x-2=0$

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

  $x^2-11x-2+{\left(\frac{11}{2}\right)}^2-{\left(\frac{11}{2}\right)}^2=0$

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

  ${\left(x-\frac{11}{2}\right)}^2-2-{\left(\frac{11}{2}\right)}^2=0$

So,

  ${\left(x-\frac{11}{2}\right)}^2-2-\frac{121}{4}=0$

So,

  ${\left(x-\frac{11}{2}\right)}^2-\left(\frac{8+121}{4}\right)=0$

Now,

  ${\left(x-\frac{11}{2}\right)}^2-\left(\frac{129}{4}\right)=0$

Now, square root of 129 is $\sqrt{129}$ and square root of 4 is 2

  ${\left(x-\frac{11}{2}\right)}^2-{\left(\frac{\sqrt{129}}{2}\right)}^2=0$

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

  $\left(x-\frac{11}{2}+\frac{\sqrt{129}}{2}\right)\left(x-\frac{11}{2}-\frac{\sqrt{129}}{2}\right)=0$

Now,

$x=\frac{11-\sqrt{129}}{2}$ or $x=\frac{11+\sqrt{129}}{2}$

Hence,

$x=-0.18$ or $x=11.18$

Nearest Tenth $x=-0.2$ or $x=11.2$