Chapter 9.4, Problem 14CYU

To determine

Find the solution.

Answer to Problem 14CYU

  $y=-\frac{1}{2}$ or $y=1$

Explanation of Solution

Given Information: $2y^2=y+1$

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

Calculation:

  $2y^2=y+1$

So,

  $2y^2-y-1=0$

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

  $y^2-\frac{y}{2}-\frac{1}{2}=0$

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

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

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

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

So,

  ${\left(y-\frac{1}{4}\right)}^2-\frac{1}{2}-\frac{1}{16}=0$

So,

  ${\left(y-\frac{1}{4}\right)}^2-\left(\frac{8+1}{16}\right)=0$

Hence,

  ${\left(y-\frac{1}{4}\right)}^2-\left(\frac{9}{16}\right)=0$

Now, square root of 9 is 3 and square root of 16 is 4.

  ${\left(y-\frac{1}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2=0$

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

  $\left(y-\frac{1}{4}+\frac{3}{4}\right)\left(y-\frac{1}{4}-\frac{3}{4}\right)=0$

So,

  $\left(y-\frac{1}{4}+\frac{3}{4}\right)=0$ or $\left(y-\frac{1}{4}-\frac{3}{4}\right)=0$

So,

  $\left(y+\frac{(-1+3)}{4}\right)=0$ or $\left(y+\frac{(-1-3)}{4}\right)=0$

Now,

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

Hence,

  $y=-\frac{1}{2}$ or $y=1$