How do you solve the equation $3 {(x - 2)}^{2} - 12 = 0$ $3(x-2)^2-12=0$

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How do you solve the equation $3 {(x - 2)}^{2} - 12 = 0$ $3(x-2)^2-12=0$

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Answer 1 · 2015-02-07T15:43:05

Lets simplify this scary thing 1st

This qill be $3 (x^{2} + 4 - 4 x) - 12 = 3 x^{2} - 12 x$ $3( x^2 + 4 - 4x) - 12 = 3x^2 - 12 x$

so this is rewritten as $3 x (x - 4) = 0$ $3x( x - 4) = 0$

The zero rule here has to be either $x - 4 = 0 or 3 x = 0$ $x- 4 = 0 or 3x = 0$

Hence the roots of these equations are $x = 4, 0$ $x =4 , 0$

Answer 2 · 2015-02-08T04:01:00

Another path you could take to solving this equation would be to pull out a GCF. As you can see, all your terms on the left are divisible by 3. And since you just have 0 on the right, you can just divide everything by 3.

Then you get:

${(x - 2)}^{2} - 4 = 0$ $(x-2)^2 - 4 = 0$

Now you could expand the binomial here, but since 4 is a perfect square, it's just much easier to move it over to the other side (add 4 to both sides), and then take the square root of both sides of the equation.

You'd get:

$x - 2 = \pm 2$ $x-2 = +-2$

Make sure you pay attention to the fact that since you took a square root, it's plus OR minus 2, and not just 2

Now, just add 2 to both sides, and you're done.

$x = 2 \pm 2 = 4 or 0$ $x = 2+-2 = 4 or 0$

Hope that helped :)

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