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

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

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Answer 1 · 2017-07-23T12:31:35

$x = \frac{\pm \sqrt{8} + 1}{4}$ $x=(+-sqrt(8)+1)/4$

Explanation:

First, we remove the $\frac{1}{2}$ $1/2$ by multiplying each side by 2 to get ${(4 x - 1)}^{2} = 8$ $(4x-1)^2=8$ .

Then to remove the square, we square root each side to get $4 x - 1 = \pm \sqrt{8}$ $4x-1=+-sqrt(8)$

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

Now, to get $x$ $x$ on its own we divide both sides by 4, $x = \frac{\pm \sqrt{8} + 1}{4}$ $x=(+-sqrt(8)+1)/4$

Answer 2 · 2017-07-23T14:56:20

$x = \frac{1}{4} + \frac{\sqrt{2}}{2}$ $color(magenta)(x=1/4+(sqrt2)/2$ or $x = 0.957106781$ $x=0.957106781$

or

$x = \frac{1}{4} - \frac{\sqrt{2}}{2}$ $color(magenta)(x=1/4-sqrt2/2$ or $x = - 0.457106781$ $x=-0.457106781$

Explanation:

$\frac{1}{2} {(4 x - 1)}^{2} = 4$ $1/2(4x-1)^2=4$

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 $x=-0.457106781$

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

2 Answers

Explanation:

Explanation:

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How do you solve \frac { 1} { 2} ( 4x - 1) ^ { 2} = 412(4x−1)2=4\frac { 1} { 2} ( 4x - 1) ^ { 2} = 4?

2 Answers

Explanation:

Explanation:

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