Chapter 9.5, Problem 8CYU

To determine

Find the solution.

Answer to Problem 8CYU

  $x=-1.4$ or $x=6.4$

Explanation of Solution

Given Information: $-2x^2+10x+22=4$

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

Calculation:

  $-2x^2+10x+22=4$

So,

  $-2x^2+10x+22-4=0$

Now, simplify

  $-2x^2+10x+18=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by -2.

  $x^2-5x-9=0$

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

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

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

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

So,

  ${\left(x-\frac{5}{2}\right)}^2-9-\frac{25}{4}=0$

So,

  ${\left(x-\frac{5}{2}\right)}^2-\left(\frac{36+25}{4}\right)=0$

Hence,

  ${\left(x-\frac{5}{2}\right)}^2-\left(\frac{61}{4}\right)=0$

Now, square root of 61 is $\sqrt{61}$ and square root of 4 is 2.

  ${\left(x-\frac{5}{2}\right)}^2-{\left(\frac{\sqrt{61}}{2}\right)}^2=0$

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

  $\left(x-\frac{5}{2}+\frac{\sqrt{61}}{2}\right)\left(x-\frac{5}{2}-\frac{\sqrt{61}}{2}\right)=0$

Now,

  $\left(x-\frac{5}{2}+\frac{\sqrt{61}}{2}\right)=0$ or $\left(x-\frac{5}{2}-\frac{\sqrt{61}}{2}\right)=0$

So,

  $\left(x-\left(\frac{5-\sqrt{61}}{2}\right)\right)=0$ or $\left(x-\left(\frac{5+\sqrt{61}}{2}\right)\right)=0$

Now,

  $x=\frac{5-\sqrt{61}}{2}$ or $x=\frac{5+\sqrt{61}}{2}$

Hence,

  $x=-1.41$ or $x=6.41$

Nearest Tenth $x=-1.4$ or $x=6.4$