Chapter 9.4, Problem 37PPS

To determine

Find the solution.

Answer to Problem 37PPS

  $b=\frac{-5}{4}$ or $b=\frac{-1}{4}$

Explanation of Solution

Given Information: $16b^2+24b+20=15$

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

Calculation:

  $16b^2+24b+20=15$

So, simplify

  $16b^2+24b+20-15=0$

Now,

  $16b^2+24b+5=0$

Now, to make coefficient of $b^2$ = 1, divide the equation by 16.

  $b^2+\frac{24b}{16}+\frac{5}{16}=0$

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

  $b^2+\frac{3b}{2}+\frac{5}{16}+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2=0$

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

  ${\left(b+\frac{3}{4}\right)}^2+\frac{5}{16}-{\left(\frac{3}{4}\right)}^2=0$

So,

  ${\left(b+\frac{3}{4}\right)}^2+\frac{5}{16}-\frac{9}{16}=0$

So,

  ${\left(b+\frac{3}{4}\right)}^2+\frac{5-9}{16}=0$

Hence,

  ${\left(b+\frac{3}{4}\right)}^2-\left(\frac{4}{16}\right)=0$

Now, square root of 4 is 2 and square root of 16 is 4.

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

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

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

Now,

  $\left(b+\frac{3}{4}+\frac{1}{2}\right)=0$ or $\left(b+\frac{3}{4}-\frac{1}{2}\right)=0$

So,

  $\left(b+\frac{3+2}{4}\right)=0$ or $\left(b+\frac{3-2}{4}\right)=0$

So,

  $\left(b+\frac{5}{4}\right)=0$ or $\left(b+\frac{1}{4}\right)=0$

Hence,

  $b=\frac{-5}{4}$ or $b=\frac{-1}{4}$