How do you solve $r^{2} - 9 r - 38 = - 9$ $r^2 - 9r - 38 = -9$ by completing the square?

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How do you solve $r^{2} - 9 r - 38 = - 9$ $r^2 - 9r - 38 = -9$ by completing the square?

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Answer 1 · 2015-05-19T22:17:46

Given $r^{2} - 9 r - 38 = - 9$ $r^2-9r-38=-9$

First add 9 to both sides to get:

$r^{2} - 9 r - 29 = 0$ $r^2-9r-29=0$

Then

$0 = r^{2} - 9 r - 29$ $0 = r^2-9r-29$

$= r^{2} - 9 r + \frac{81}{4} - \frac{81}{4} - 29$ $= r^2-9r+81/4-81/4-29$

$= {(r - \frac{9}{2})}^{2} - (\frac{81}{4} + 29)$ $= (r-9/2)^2-(81/4+29)$

$= {(r - \frac{9}{2})}^{2} - (\frac{81}{4} + \frac{116}{4})$ $= (r-9/2)^2-(81/4+116/4)$

$= {(r - \frac{9}{2})}^{2} - \frac{197}{4}$ $= (r-9/2)^2-197/4$

Adding $\frac{197}{4}$ $197/4$ to both ends of this equation, we find:

${(r - \frac{9}{2})}^{2} = \frac{197}{4}$ $(r-9/2)^2 = 197/4$

So:

$r - \frac{9}{2} = \pm \sqrt{\frac{197}{4}} = \pm \frac{\sqrt{197}}{\sqrt{4}} = \pm \frac{\sqrt{197}}{2}$ $r-9/2 = +-sqrt(197/4) = +-sqrt(197)/sqrt(4) = +-sqrt(197)/2$

Add $\frac{9}{2}$ $9/2$ to both sides to get:

$r = \frac{9}{2} \pm \frac{\sqrt{197}}{2} = \frac{9 \pm \sqrt{197}}{2}$ $r = 9/2+-sqrt(197)/2 = (9+-sqrt(197))/2$

In general:

$a r^{2} + b r + c = a {(r + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ar^2+br+c = a(r+b/(2a))^2 + (c - b^2/(4a))$

which is zero when

$a {(r + \frac{b}{2 a})}^{2} = (\frac{b^{2}}{4 a} - c) = \frac{b^{2} - 4 a c}{4 a}$ $a(r+b/(2a))^2 = (b^2/(4a) - c) = (b^2 - 4ac)/(4a)$

and hence

${(r + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$ $(r+b/(2a))^2 = (b^2 - 4ac)/(4a^2)$

so

$r + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} = \pm \frac{\sqrt{b^{2} - 4 a c}}{\sqrt{4 a^{2}}}$ $r+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)$

$= \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $= +-sqrt(b^2-4ac)/(2a)$

Subtracting $\frac{b}{2 a}$ $b/(2a)$ from both sides we get:

$r = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $r = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b +- sqrt(b^2-4ac))/(2a)$

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How do you solve $r^{2} - 9 r - 38 = - 9$ $r^2 - 9r - 38 = -9$ by completing the square?

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How do you solve r2−9r−38=−9r^2 - 9r - 38 = -9 by completing the square?

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How do you solve $r^{2} - 9 r - 38 = - 9$ $r^2 - 9r - 38 = -9$ by completing the square?