sqrt(x+8)^2 = sqrt(81)
x+8 =9
x=1
2007-12-26 04:22:19
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answer #1
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answered by WhatWasThatNameAgain? 5
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x+8)² = 81
If we notice|fix (x+8)², of a remarkable product or also enumenat remarkable formula, which he|she|it says: The square of the sum of two numbers him equal to|in the square of the first, more the double of
frequents the first for the second, more the square of the second.
Therefore:
x² + 2x(8) + 8² = 81
x² + 16x +64 = 81
x² + 16x + 64 - 81= 0
x² + 16x - 17= 0
This qu once solved by the usual procedure is a quadratic equation,
it faces|gives:
x' = 1
x" = -17
Greetings from Catalonia and until the next one
Good Christmas
2007-12-26 11:24:13
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answer #2
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answered by josep d 7
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(x+8)² = 81
x + 8 = â81
x + 8 = ±9
Remember for a quadratic equation there are two roots, since the square root of 81 is positive or negative 9 you have two real solutions.
x = -9 -8 = -17
x = 9 -8 = 1
2007-12-26 04:26:50
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answer #3
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answered by Brian K² 6
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Step one: Take the square root of both sides. You will get:
x+8=9 OR x+8=-9, because 9^2 is 81, and so is (-9)^2.
Step two: Subtract 8 from both sides.
You get x=1 or x=-17
Hope I helped!!
:o)
2007-12-26 04:28:03
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answer #4
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answered by :o) Sunshine 3
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x + 8 = ± 9
x = 1 , x = - 17
2007-12-29 19:53:39
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answer #5
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answered by Como 7
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(x + 8)^2 = 81
x + 8 = â81
x + 8 = 9
x = 9 - 8
x = 1
2007-12-26 04:23:07
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answer #6
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answered by An ESL Learner 7
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x^2+2*x*8+64=81
x^2+16*x=17
x^2+16*x-17=0
D=b^2-4*a*g
=256-4*1*(-17)
=256-68
=188>0
X 1,2= -b + / - D
= -16 + / - (root of number 188)
2007-12-26 04:34:55
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answer #7
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answered by P Defere 1
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(x+8)^2 = 81
taking sq rt from both side
x+8 = 9 or x+8 = -9
x = 1 or x = -17
2007-12-26 04:25:24
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answer #8
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answered by jaya 4
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(x+8)*2=9*2
x=1
2007-12-26 05:48:06
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answer #9
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answered by praman18 2
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x^2+16x+64=81
x^2+16x-17=0
x^2+17x-x-17=0
x(x+17)-1(x+17)=0
(x+17)(x-1)=0
x=1 or x=-17
2007-12-26 04:27:48
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answer #10
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answered by cidyah 7
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