Chapter 9.5, Problem 36PPS

To determine

Find the solution.

Answer to Problem 36PPS

$x=-0.5$ or $x=2.5$

Explanation of Solution

Given Information: $\frac{2}{3}{x}^2-\frac{4}{3}x=\frac{5}{6}$

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

Calculation:

  $\frac{2}{3}{x}^2-\frac{4}{3}x=\frac{5}{6}$

So, simplify

  $\frac{2}{3}{x}^2-\frac{4}{3}x-\frac{5}{6}=0$

Now, to make coefficient of $x^2$ = 1, multiply the equation by $\frac{3}{2}$ .

  $x^2-2x-\frac{5}{4}=0$

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

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

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

  ${\left(x-1\right)}^2-\frac{5}{4}-{\left(1\right)}^2=0$

So,

  ${\left(x-1\right)}^2-\left(\frac{5+4}{4}\right)=0$

Now,

  ${\left(x-1\right)}^2-\left(\frac{9}{4}\right)=0$

Now, square root of 9 is 3 and square root of 4 is 2.

  ${\left(x-1\right)}^2-{\left(\frac{3}{2}\right)}^2=0$

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

  $\left(x-1+\frac{3}{2}\right)\left(x-1-\frac{3}{2}\right)=0$

Now,

$x=1-\frac{3}{2}$ or $x=1+\frac{3}{2}$

Now,

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

Now,

$x=-\frac{1}{2}$ or $x=\frac{5}{2}$

Hence,

$x=-0.5$ or $x=2.5$