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

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

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Answer 1 · 2017-01-24T20:54:13

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

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 $7 \cdot x = \frac{84}{2}$ $7 * x = 84/2$ , so that when you expand the bracket, you get two terms of $42 b$ $42b$ which added together gives $84 b$ $84b$ . $7 \cdot 6 = 42$ $7 * 6 = 42$ , 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 $6 \cdot 6 = 36$ $6 * 6 = 36$ , 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 $2^{2}$ $2^2$ and ${(- 2)}^{2}$ $(-2)^2$ give 4, you arrive at 2 different answers, both of which are valid.

Answer 2 · 2017-01-24T20:58:53

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

Explanation:

The difference of squares identity can be written:

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

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

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

So we find:

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

$0 = {(7 b)}^{2} + 2 (7 b) (6) + {(6)}^{2} - 4$ $color(white)(0) = (7b)^2+2(7b)(6)+(6)^2-4$

$0 = {(7 b)}^{2} + 2 (7 b) (6) + {(6)}^{2} - 4$ $color(white)(0) = (7b)^2+2(7b)(6)+(6)^2-4$

$0 = {(7 b + 6)}^{2} - 2^{2}$ $color(white)(0) = (7b+6)^2-2^2$

$0 = ((7 b + 6) - 2) ((7 b + 6) + 2)$ $color(white)(0) = ((7b+6)-2)((7b+6)+2)$

$0 = (7 b + 4) (7 b + 8)$ $color(white)(0) = (7b+4)(7b+8)$

Hence:

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

Answer 3 · 2017-01-25T00:15:25

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

Explanation:

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

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

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

So for this particular problem we have

Step 1: Divide by $a = 49$ $a=49$

$49 b^{2} + 84 b + 32 = 0$ $49b^2 + 84b + 32 = 0$
$:. b^2 + 84/49b + 32/49 = 0$
$:. b^2 + 12/7b + 32/49 = 0$

Step 2: Form a perfect square using $(x+b/2)^2$

$:. (b + 1/2*12/7)^2 - (1/2*12/7)^2 + 32/49=0$
$:. (b + 6/7)^2 - (6/7)^2 + 32/49=0$
$:. (b + 6/7)^2 - 36/49 + 32/49=0$
$:. (b + 6/7)^2 - 4/49=0$

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

$:. (b + 6/7)^2 = 4/49$
$:. b + 6/7 = +-sqrt(4/49)$
$:. b + 6/7 = +-2/7$
$:. b = - 6/7 +-2/7$

Leading to the two solutions:

$b = - 6/7 -2/7 = -8/7$

$b = - 6/7 +2/7 = -4/7$

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

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How do you solve 49b^2 + 84b + 32 = 0 49b2+84b+32=049b^2 + 84b + 32 = 0 by completing the square?

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