Chapter 9.6, Problem 2BGP

To determine

Find the solution.

Answer to Problem 2BGP

  $x=-1.43$ or $x=2.1$

Explanation of Solution

Given Information: $3x^2-2x-9=0$

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

Calculation:

  $3x^2-2x-9=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 3.

  $x^2-\frac{2}{3}x-\frac{9}{3}=0$

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

  $x^2-\frac{2}{3}x-3+{\left(\frac{2}{6}\right)}^2-{\left(\frac{2}{6}\right)}^2=0$

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

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

So,

  ${\left(x-\frac{1}{3}\right)}^2-3-\frac{1}{9}=0$

Now,

  ${\left(x-\frac{1}{3}\right)}^2-\frac{27-1}{9}=0$

Now,

  ${\left(x-\frac{1}{3}\right)}^2-\frac{28}{9}=0$

Now, square root of 28 is $\sqrt{28}$ and square root of 9 is 3

  ${\left(x-\frac{1}{3}\right)}^2-{\left(\frac{\sqrt{28}}{3}\right)}^2=0$

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

  $\left(x-\frac{1}{3}+\frac{\sqrt{28}}{3}\right)\left(x-\frac{1}{3}-\frac{\sqrt{28}}{3}\right)=0$

Now,

  $x=\frac{1}{3}-\frac{\sqrt{28}}{3}$ or $x=\frac{1}{3}+\frac{\sqrt{28}}{3}$

Now,

  $x=\frac{1-\sqrt{28}}{3}$ or $x=\frac{1+\sqrt{28}}{3}$

Hence,

  $x=-1.43$ or $x=2.1$