Chapter 9.6, Problem 4AGP

To determine

Find the solution.

Answer to Problem 4AGP

$x=-3$ or $x=-2.5$

Explanation of Solution

Given Information: $2x^2+11x+15=0$

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

Calculation:

  $2x^2+11x+15=0$

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

  $x^2+\frac{11}{2}x+\frac{15}{2}=0$

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

  $x^2+\frac{11}{2}x+\frac{15}{2}+{\left(\frac{11}{4}\right)}^2-{\left(\frac{11}{4}\right)}^2=0$

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

  ${\left(x+\frac{11}{4}\right)}^2+\frac{15}{2}-{\left(\frac{11}{4}\right)}^2=0$

So,

  ${\left(x+\frac{11}{4}\right)}^2+\frac{15}{2}-\frac{121}{16}=0$

Now,

  ${\left(x+\frac{11}{4}\right)}^2+\frac{120-121}{16}=0$

Now,

  ${\left(x+\frac{11}{4}\right)}^2-\frac{1}{16}=0$

Now, square root of 1 is 1 and square root of 16 is 4

  ${\left(x+\frac{11}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2=0$

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

  $\left(x+\frac{11}{4}+\frac{1}{4}\right)\left(x+\frac{11}{4}-\frac{1}{4}\right)=0$

Now,

$x=-\frac{11}{4}-\frac{1}{4}$ or $x=-\frac{11}{4}+\frac{1}{4}$

Now,

$x=\frac{-11-1}{4}$ or $x=\frac{-11+1}{4}$

Now,

$x=-\frac{12}{4}$ or $x=\frac{-10}{4}$

Hence,

$x=-3$ or $x=-2.5$