Question

How do you solve `9x^2-4=0`?

Answer 1

`x=\pm\frac{2}{3}`

Explanation:

`9x^2-4=0`

Move the `4` to the RHS

`9x^2=4`

Divide both sides by `9`

`x^2=\frac{4}{9}`

Take the square root of both sides

`\sqrt{x^2}=\pm\sqrt{\frac{4}{9}}`

Simplify

`x=\pm\frac{2}{3}`

`\thereforex=-\frac{2}{3},\qquad\qquad\qquad x=\frac{2}{3}`

Answer 2

`x= +-sqrt(4/9`

In decimal `x = +-0.67`

Explanation:

`9x^2 - 4=0`
Let start by adding `color(red)(4)` to both sides
`9x^2 cancel(- 4) cancelcolor(red)(+4)=0+color(red)(4)`
`9x^2 = 4`
Divide both sides by `color(green)(9)`
`(cancel9x^2)/cancelcolor(green)(9) = 4/color(green)(9)`
`x^2 = 4/9`
We need to take square root `sqrt`
`x= +-sqrt(4/9)`