Chapter 9.6, Problem 1AGP

To determine

Find the solution.

Answer to Problem 1AGP

  $x=-6$ or $x=1.5$

Explanation of Solution

Given Information: $2x^2+9x=18$

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

Calculation:

  $2x^2+9x=18$

So, simplify

  $2x^2+9x-18=0$

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

  $x^2+\frac{9}{2}x-\frac{18}{2}=0$

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

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

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

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

So,

  ${\left(x+\frac{9}{4}\right)}^2-9-\frac{81}{16}=0$

Now,

  ${\left(x+\frac{9}{4}\right)}^2-\left(\frac{144+81}{16}\right)=0$

Now,

  ${\left(x+\frac{9}{4}\right)}^2-\frac{225}{16}=0$

Now, square root of 225 is 15 and square root of 16 is 4

  ${\left(x+\frac{9}{4}\right)}^2-{\left(\frac{15}{4}\right)}^2=0$

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

  $\left(x+\frac{9}{4}+\frac{15}{4}\right)\left(x+\frac{9}{4}-\frac{15}{4}\right)=0$

Now,

  $x=-\frac{9}{4}-\frac{15}{4}$ or $x=-\frac{9}{4}+\frac{15}{4}$

Now,

  $x=\frac{-9-15}{4}$ or $x=\frac{-9+15}{4}$

Now,

  $x=-\frac{24}{4}$ or $x=\frac{6}{4}$

Hence,

  $x=-6$ or $x=1.5$