Question

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

Answer 1

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`