Question

How do you solve `2x^2-5=0`?

Answer 1

`x_(1,2) = +- sqrt(10)/2`

Explanation:

You need to take three steps in order to solve this equation

• add `5` to both sides of the equation

`2x^2 - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 0 + 5`

`2x^2 = 5`

• divide both sides of the equation by `2`

`(color(red)(cancel(color(black)(2)))x^2)/color(red)(cancel(color(black)(2))) = 5/2`

`x^2 = 5/2`

• take the square root of both sides to solve for `x`

`sqrt(x^2) = sqrt(5/2)`

`x_(1,2) = +- sqrt(5)/sqrt(2)`

You can simplify this further by rationalizing the denominator if the fraction. To do that, multiply the fraction by `1 = sqrt(2)/sqrt(2)`

`x_(1,2) = +- (sqrt(5) * sqrt(2))/(sqrt(2) * sqrt(2))`

`x_(1,2) = +- sqrt(10)/2`

This means that your equation has two solutions,

`x_1 = color(green)(sqrt(10)/2)" "` or `" "x_2 = color(green)(-sqrt(10)/2)`