Chapter 9.4, Problem 40PPS

To determine

Find the solution.

Answer to Problem 40PPS

  $y=\frac{-8}{7}$ or $y=1$

Explanation of Solution

Given Information: $14y^2=-2y+16$

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

Calculation:

  $14y^2=-2y+16$

So, simplify

  $14y^2+2y-16=0$

Now, to make coefficient of $y^2$ = 1, divide the equation by 14.

  $y^2+\frac{2y}{14}-\frac{16}{14}=0$

Now,

  $y^2+\frac{y}{7}-\frac{8}{7}=0$

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

  $y^2+\frac{y}{7}-\frac{8}{7}+{\left(\frac{1}{14}\right)}^2-{\left(\frac{1}{14}\right)}^2=0$

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

  ${\left(y+\frac{1}{14}\right)}^2-\frac{8}{7}-{\left(\frac{1}{14}\right)}^2=0$

So,

  ${\left(y+\frac{1}{14}\right)}^2-\frac{8}{7}-\frac{1}{196}=0$

So,

  ${\left(y+\frac{1}{14}\right)}^2-\left(\frac{224+1}{196}\right)=0$

Hence,

  ${\left(y+\frac{1}{14}\right)}^2-\left(\frac{225}{196}\right)=0$

Now, square root of 225 is 15 and square root of 196 is 14.

  ${\left(y+\frac{1}{14}\right)}^2-{\left(\frac{15}{14}\right)}^2=0$

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

  $\left(y+\frac{1}{14}+\frac{15}{14}\right)\left(y+\frac{1}{14}-\frac{15}{14}\right)=0$

Now,

  $\left(y+\frac{1}{14}+\frac{15}{14}\right)=0$ or $\left(y+\frac{1}{14}-\frac{15}{14}\right)=0$

So,

  $\left(y+\frac{1+15}{14}\right)=0$ or $\left(x+\frac{1-15}{14}\right)=0$

So,

  $\left(y+\frac{16}{14}\right)=0$ or $\left(x-\frac{14}{14}\right)=0$

Now, simplify

  $y=\frac{-16}{14}$ or $y=\frac{14}{14}$

Hence,

  $y=\frac{-8}{7}$ or $y=1$