# Math- Algebra

This one showed up in the middle of my quadratic equations. There is no y or 0. What am I supposed to do with it?

Give exact and approximate solutions to three decimal places.

(x+5)^2=81

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1. x + 5 = + sqrt 81 = +9
or
x + 5 = - sqrt 81 = -9

so
x = 9-5 = 4
or
x = -9-5 = -14
check those back

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2. How to check back:
if x = 4
what is (x+5)^2 ? hint 9^2 = 81 !!!

if x = -9
what is (x+5)^2 ? hint (-9)*(-9) = 81
we did it :)

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3. perhaps it is coffee time

if x = -14
what is (x+5)^2 ? hint (-9)*(-9) = 81
we did it :)

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4. Oh, and to do this as quadratic equation:
(x+5)^2 = 81
x^2 + 10 x + 25 = 81

x^2 + 10 x - 56 = 0

x = -10/2 +/- (1/2) sqrt(100 + 4*1*56)

x = -5 +/- (1/2)sqrt ( 324)

sqrt 324 = 18 by the way

x = -5 +/- 9

x = -5+9 = +4
or
x = -5-9 = -14

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5. You catching on to this ?

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6. LOL. Your great, thanks.

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7. Yeah, well check the one below. I did not read carefully and left the wall of the barn out !!

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