Chapter 9.4, Problem 12CYU

To determine

Find the solution.

Answer to Problem 12CYU

$b=-10$ or $b=3$

Explanation of Solution

Given Information: $b^2+7b-30=0$

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

Calculation:

  $b^2+7b-30=0$

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

  $b^2+7b-30+{\left(\frac{7}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2=0$

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

  ${\left(b+\frac{7}{2}\right)}^2-30-{\left(\frac{7}{2}\right)}^2=0$

So,

  ${\left(b+\frac{7}{2}\right)}^2-30-\frac{49}{4}=0$

So,

  ${\left(b+\frac{7}{2}\right)}^2-\left(\frac{120+49}{4}\right)=0$

Hence,

  ${\left(b+\frac{7}{2}\right)}^2-\left(\frac{169}{4}\right)=0$

Now, square root of 25 is 5 and square root of 16 is 4.

  ${\left(b+\frac{7}{2}\right)}^2-{\left(\frac{13}{2}\right)}^2=0$

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

  $\left(b+\frac{7}{2}+\frac{13}{2}\right)\left(b+\frac{7}{2}-\frac{13}{2}\right)=0$

Now,

$\left(b+\frac{7}{2}+\frac{13}{2}\right)=0$ or $\left(b+\frac{7}{2}-\frac{13}{2}\right)=0$

So,

$\left(b+\frac{7+13}{2}\right)=0$ or $\left(b+\frac{7-13}{2}\right)=0$

So,

$\left(b+\frac{20}{2}\right)=0$ or $\left(b-\frac{6}{2}\right)=0$

Now,

$b=\frac{-20}{2}$ or $b=\frac{6}{2}$

Hence,

$b=-10$ or $b=3$