Chapter 9.4, Problem 13CYU

To determine

Find the solution.

Answer to Problem 13CYU

  $x=3.57$ or $x=-1.17$

Explanation of Solution

Given Information: $5x^2-12x-7=14$

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

Calculation:

  $5x^2-12x-7=14$

So,

  $5x^2-12x-7-14=0$

Now, simplify

  $5x^2-12x-21=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 5.

  $x^2-\frac{12}{5}x-\frac{21}{5}=0$

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

  $x^2-\frac{12}{5}x-\frac{21}{5}+{\left(\frac{12}{10}\right)}^2-{\left(\frac{12}{10}\right)}^2=0$

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

  ${\left(x-\frac{6}{5}\right)}^2-\frac{21}{5}-{\left(\frac{6}{5}\right)}^2=0$

So,

  ${\left(x-\frac{6}{5}\right)}^2-\frac{21}{5}-\frac{36}{25}=0$

So,

  ${\left(x-\frac{6}{5}\right)}^2-\left(\frac{105+36}{25}\right)=0$

Hence,

  ${\left(x-\frac{6}{5}\right)}^2-\frac{141}{25}=0$

Now, square root of 141 is $\sqrt{141}$ and square root of 25 is 5.

  ${\left(x-\frac{6}{5}\right)}^2-{\left(\frac{\sqrt{141}}{5}\right)}^2=0$

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

  $\left(x-\frac{6}{5}+\frac{\sqrt{141}}{5}\right)\left(x-\frac{6}{5}-\frac{\sqrt{141}}{5}\right)=0$

Now,

  $\left(x-\frac{6}{5}+\frac{\sqrt{141}}{5}\right)=0$ or $\left(x-\frac{6}{5}-\frac{\sqrt{141}}{5}\right)=0$

So,

  $\left(x-\frac{6+\sqrt{141}}{5}\right)=0$ or $\left(x-\frac{6-\sqrt{141}}{5}\right)=0$

Now,

  $x=\frac{6+\sqrt{141}}{5}$ or $x=\frac{6-\sqrt{141}}{5}$

Hence,

  $x=3.57$ or $x=-1.17$