How do you solve #\frac { 1} { 2} ( 4x - 1) ^ { 2} = 4#?

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Answer 1 · 2017-07-23T12:31:35

#x=(+-sqrt(8)+1)/4#

First, we remove the #1/2# by multiplying each side by 2 to get #(4x-1)^2=8#.

Then to remove the square, we square root each side to get #4x-1=+-sqrt(8)#

We just need the#4x# on its own, so we add 1 to each side, #4x=+-sqrt(8)+1#.

Now, to get #x# on its own we divide both sides by 4, #x=(+-sqrt(8)+1)/4#

Answer 2 · 2017-07-23T14:56:20

#color(magenta)(x=1/4+(sqrt2)/2# or #x=0.957106781#

or

#color(magenta)(x=1/4-sqrt2/2# or #x=-0.457106781#

#1/2(4x-1)^2=4#

Multiply both sides by 2

#:.(4x-1)^2=8#

#:.4x-1=sqrt8#

#:.4x=sqrt8+1#

#:.x=(sqrt8+1)/4#

#:.x=(sqrt(2*2*2)+1)/4#

#:.sqrt2*sqrt2=2#

#x=(2sqrt2+1)/4#

#x=1/4+(cancel2^1sqrt2)/cancel4^2#

#x=1/4+-sqrt2/2#

#color(magenta)(x=1/4+sqrt2/2# or #x=0.957106781#

#color(magenta)(x=1/4-sqrt2/2# or #x=-0.457106781#