How do you solve $3 x^{2} - 8 = 4$ $3x^2 - 8 = 4$ ?

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Answer 1 · 2016-06-19T01:31:28

Set the expression equal to 0, then simplify and solve to get to $x = 2, - 2$ $x=2, -2$

Let's start with the original:

$3 x^{2} - 8 = 4$ $3x^2-8=4$

Subtract 4 from both sides to set the right side of the equation to 0 and we have:

$3 x^{2} - 12 = 0$ $3x^2-12=0$

We can then divide both sides by 3:

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

And factor that:

$(x + 2) (x - 2) = 0$ $(x+2)(x-2)=0$

Now let's solve each one:

$x + 2 = 0$ $x+2=0$
$x = - 2$ $x=-2$

$x - 2 = 0$ $x-2=0$
$x = 2$ $x=2$

And so $x = 2, - 2$ $x=2, -2$

How do you solve 3x2−8=43x^2 - 8 = 4?