How do you solve $2 {(x - 3)}^{2} = 8$ $2(x-3)^2=8$ ?

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How do you solve $2 {(x - 3)}^{2} = 8$ $2(x-3)^2=8$ ?

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Answer 1 · 2017-07-11T21:59:23

$x = 5$ $x=5$ or $x = 1$ $x=1$

See below

Explanation:

$2 {(x - 3)}^{2} = 8$ $2(x−3)^2=8$

Divide both sides by 2: ${(x - 3)}^{2} = 4$ $(x-3)^2=4$

Take the square root of each side:

$(x - 3) = \pm 2$ $(x-3)=+-2$

$x - 3 = + 2 \to x = 5$ $x-3=+2" "rarrx=5$
or
$x - 3 = - 2 \to x = 1$ $x-3=-2" "rarrx=1$

$x = 5$ $x=5$ or $x = 1$ $x=1$

Answer 2 · 2017-07-11T22:04:52

The answer is x= 5 and 1

Explanation:

First you start by evualting the $2 {(x - 3)}^{2}$ $2(x-3)^2$ part of the equation.
$2 {(x - 3)}^{2}$ $2(x-3)^2$ =8
$2 (x^{2} - 6 x + 9) = 8$ $2(x^2-6x+9)=8$
$2 x^{2} - 12 x + 18 = 8$ $2x^2-12x+18=8$
Then you move over the constant on the right side to the left .
$2 x^{2} - 12 x + 18 - 8 = 0$ $2x^2-12x+18-8=0$
$2 x^{2} - 12 x + 10 = 0$ $2x^2-12x+10=0$
Then divide the whole equation by 2.
$\frac{2 x^{2} - 12 x + 10}{2} = \frac{0}{2}$ $(2x^2-12x+10)/2=0/2$
$x^{2} - 6 x + 5 = 0$ $x^2-6x+5=0$
Then factor the equation normally.
$(x - 5) (x - 1) = 0$ $(x-5)(x-1)=0$

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How do you solve $2 {(x - 3)}^{2} = 8$ $2(x-3)^2=8$ ?

2 Answers

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