How do you find the solution to the quadratic equation $3x^2-8=0$ ?

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Answer 1 · 2015-05-13T06:56:23

Transposing $8$ to the other side, we get:

$3x^2 = 8$

Dividing both sides by 3, we get:

$(cancel(3)x^2)/cancel(3) = 8/3$

$x^2 = 8/3$

Taking square root on both sides gives us:

$sqrt(x^2) = sqrt(8/3)$

$x = +-sqrt8/sqrt3$

$x = +-(2sqrt2)/sqrt3$

To rationalise the denominator,

$x = +- ((2sqrt2)/sqrt3)*(sqrt3/sqrt3)$

$x = +- (2sqrt2*sqrt3)/3$

We know that $color(blue)(sqrta*sqrtb = sqrt(ab)$

Hence $x = +- (2sqrt(2*3))/3$

$color(green)( x = (2sqrt6)/3,-(2sqrt6)/3$

How do you find the solution to the quadratic equation 3x^2-8=03x^2-8=0?