# How do you solve 49b^2 + 84b + 32 = 0  by completing the square?

Jan 24, 2017

${\left(7 b + 6\right)}^{2} + 32 - 36 = 0$
${\left(7 b + 6\right)}^{2} = 4$
$7 b + 6 = \pm 2$
$7 b = - 4$ or $7 b = - 8$
$b = - \frac{4}{7} , - \frac{8}{7}$

#### Explanation:

The first step is to find the number that squares to give 49, which is 7. Then, you need to find a number such that $7 \cdot x = \frac{84}{2}$, so that when you expand the bracket, you get two terms of $42 b$ which added together gives $84 b$. $7 \cdot 6 = 42$, so 6 is the number we want. Now, putting that into the bracket works fine, but you have to account for the fact that when expanded it will give an extra $6 \cdot 6 = 36$, so subtract 36 outside the bracket.

After that, it's a case of square rooting both sides, then rearranging to get the answer. Since both ${2}^{2}$ and ${\left(- 2\right)}^{2}$ give 4, you arrive at 2 different answers, both of which are valid.

Jan 24, 2017

$b = - \frac{4}{7} \text{ }$ or $\text{ } b = - \frac{8}{7}$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We will use this with $A = 7 b$ and $B = 6$...

Note that $49 = {7}^{2}$, $\frac{84}{2 \cdot 7} = 6$ and ${6}^{2} = 36$

So we find:

$0 = 49 {b}^{2} + 84 b + 32$

$\textcolor{w h i t e}{0} = {\left(7 b\right)}^{2} + 2 \left(7 b\right) \left(6\right) + {\left(6\right)}^{2} - 4$

$\textcolor{w h i t e}{0} = {\left(7 b\right)}^{2} + 2 \left(7 b\right) \left(6\right) + {\left(6\right)}^{2} - 4$

$\textcolor{w h i t e}{0} = {\left(7 b + 6\right)}^{2} - {2}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(7 b + 6\right) - 2\right) \left(\left(7 b + 6\right) + 2\right)$

$\textcolor{w h i t e}{0} = \left(7 b + 4\right) \left(7 b + 8\right)$

Hence:

$b = - \frac{4}{7} \text{ }$ or $\text{ } b = - \frac{8}{7}$

Jan 25, 2017

$b = - \frac{8}{7} , - \frac{4}{7}$

#### Explanation:

This is the method I was taught to complete the square on a general quadratic:

${x}^{2} + b x + c$

• Step 1: Factor out (or divide) the coefficient of ${x}^{2}$ so that that coefficient is $1$.
• Step 2:Use the knowledge of a perfect square ${\left(x + \alpha\right)}^{2} = {x}^{2} + 2 \alpha x + {\alpha}^{2}$, Here we have $2 \alpha = b \implies \alpha = \frac{1}{2} b$ and subtract ${\alpha}^{2} = {\left(\frac{1}{2} b\right)}^{2}$
• Step 3: Solve the equation

So for this particular problem we have

Step 1: Divide by $a = 49$

$49 {b}^{2} + 84 b + 32 = 0$
$\therefore {b}^{2} + \frac{84}{49} b + \frac{32}{49} = 0$
$\therefore {b}^{2} + \frac{12}{7} b + \frac{32}{49} = 0$

Step 2: Form a perfect square using ${\left(x + \frac{b}{2}\right)}^{2}$

$\therefore {\left(b + \frac{1}{2} \cdot \frac{12}{7}\right)}^{2} - {\left(\frac{1}{2} \cdot \frac{12}{7}\right)}^{2} + \frac{32}{49} = 0$
$\therefore {\left(b + \frac{6}{7}\right)}^{2} - {\left(\frac{6}{7}\right)}^{2} + \frac{32}{49} = 0$
$\therefore {\left(b + \frac{6}{7}\right)}^{2} - \frac{36}{49} + \frac{32}{49} = 0$
$\therefore {\left(b + \frac{6}{7}\right)}^{2} - \frac{4}{49} = 0$

Step 3:: If we are solving an equation then solve it

$\therefore {\left(b + \frac{6}{7}\right)}^{2} = \frac{4}{49}$
$\therefore b + \frac{6}{7} = \pm \sqrt{\frac{4}{49}}$
$\therefore b + \frac{6}{7} = \pm \frac{2}{7}$
$\therefore b = - \frac{6}{7} \pm \frac{2}{7}$

$b = - \frac{6}{7} - \frac{2}{7} = - \frac{8}{7}$
$b = - \frac{6}{7} + \frac{2}{7} = - \frac{4}{7}$