SOLUTION: How do I "simplify" The square root of 1/32? (that is 1/32 inside the radical.) I have a complicated question and can get so far but I forget how to do this step. The problem sta

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Question 51362This question is from textbook college algebra
: How do I "simplify" The square root of 1/32? (that is 1/32 inside the radical.)
I have a complicated question and can get so far but I forget how to do this step. The problem starts as 4x^2+x-1/16=0 I'm suppose to use the "complete the square method" HELP!! This question is from textbook college algebra

Found 2 solutions by THANApHD, stanbon:
Answer by THANApHD(104) (Show Source):
You can put this solution on YOUR website!
(a+b)^2 = a^2+2ab+b^2

4x^2+x-1/16 = 0
x^2+x/4 = 1/64
x^2+x/4+1/64= 1/64+1/64
(x+1/8)^2 = 1/32
x = (1/32)^(1/2)-1/8
= 1/(16*2)^(1/2)-1/8
= 1/4*(2)^(1/2)-1/8
= 2^(1/2)/8-1/8
= [2^(1/2)-1]/8
= [2^(1/2)-1]/8 or x = {-[2^(1/2)]-1}/8
= (1.414.. - 1)/8 or x = (-1.414.. -1)/8
= 0.414/8 or x = -2.414/8
= 0.0517... or x = -0.3017

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
How do I "simplify" The square root of 1/32? (that is 1/32 inside the radical.)
sqrt(1/32)= [sqrt1] /[ sqrt32]= [sqrt2]/[sqrt64]=[sqrt2]/8
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I have a complicated question and can get so far but I forget how to do this step. The problem starts as 4x^2+x-1/16=0
Rewrite as:
4(x^2+(1/4)x+?=(1/16+?
Complete the square, as follows:
4(x^2+(1/4)x+(1/8)^2)=(1/16)+4(1/8)^2
4(x+(1/8))^2= 17/16
(x+(1/8))^2=17/64
Take the square root of both sides to get:
x+1/8 = [sqrt17]/8 or x+1/8 = sqrt 17]/8-
x=[-1+sqrt 17]/8 or x=[-1-sqrt 17]/8
Cheers,
Stan H.