Chapter 9.4, Problem 39PPS

To determine

Find the solution.

Answer to Problem 39PPS

  $x=\frac{-3}{4}$ or $x=\frac{-2}{3}$

Explanation of Solution

Given Information: $48x^2+68x+24=0$

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

Calculation:

  $48x^2+68x+24=0$

So, simplify

  $12x^2+17x+6=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 12.

  $x^2+\frac{17x}{12}+\frac{6}{12}=0$

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

  $x^2+\frac{17x}{12}+\frac{1}{2}+{\left(\frac{17}{24}\right)}^2-{\left(\frac{17}{24}\right)}^2=0$

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

  ${\left(x+\frac{17}{24}\right)}^2+\frac{1}{2}-{\left(\frac{17}{24}\right)}^2=0$

So,

  ${\left(x+\frac{17}{24}\right)}^2+\frac{1}{2}-\frac{289}{576}=0$

So,

  ${\left(x+\frac{17}{24}\right)}^2+\frac{288-289}{576}=0$

Hence,

  ${\left(x+\frac{17}{24}\right)}^2-\left(\frac{1}{576}\right)=0$

Now, square root of 1 is 1 and square root of 576 is 24.

  ${\left(x+\frac{17}{24}\right)}^2-{\left(\frac{1}{24}\right)}^2=0$

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

  $\left(x+\frac{17}{24}+\frac{1}{24}\right)\left(x+\frac{17}{24}-\frac{1}{24}\right)=0$

Now,

  $\left(x+\frac{17}{24}+\frac{1}{24}\right)=0$ or $\left(x+\frac{17}{24}-\frac{1}{24}\right)=0$

So,

  $\left(x+\frac{17+1}{24}\right)=0$ or $\left(x+\frac{17-1}{24}\right)=0$

So,

  $\left(x+\frac{18}{24}\right)=0$ or $\left(x+\frac{16}{24}\right)=0$

Now, simplify

  $x=\frac{-18}{24}$ or $x=\frac{-16}{24}$

Hence,

  $x=\frac{-3}{4}$ or $x=\frac{-2}{3}$