Question

Solve the following equation by factoring:9x^2-3x-2=0

Answer 1

The two roots of the quadratic equation are

`x_1= - (1)/(3) \text{ and } x_2= (2)/(3)`

Explanation:

Original quadratic equation is
`9x^(2)-3x-2=0`

Divide both sides by 9:

`x^(2) - (x)/(3) - (2)/(9)=0`

Add
`(2)/(9)` to both sides to get rid of the constant on the LHS

`x^(2) - (x)/(3) - (2)/(9)+(2)/(9)=(2)/(9)` ==>
`x^(2) - (x)/(3)=(2)/(9)`

Add
`(1)/(36)` to both sides

`x^(2) - (x)/(3)+(1)/(36)=(2)/(9) +(1)/(36)`

This simplifies to

`x^(2) - (x)/(3)+(1)/(36)=(1)/(4)`

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b =
`(1)/(6)\right)` we can see that

`\left(x - (1)/(6)\right)^2` =
`x^2 - 2.x. (-(1)/(6)) + (1)/(36) = x^(2) - (x)/(3)+(1)/(36)`

So

`\left(x - (1)/(6)\right)^2=(1)/(4)`

Taking square roots on both sides

`\left(x - (1)/(6)\right)^2= \pm(1)/(4)`

So the two roots or solutions of the equation are

`x - (1)/(6)=-\sqrt{(1)/(4)}` and
`x - (1)/(6)=\sqrt{(1)/(4)}`

`\sqrt{(1)/(4)} = (1)/(2)`

So the two roots are

`x_1=(1)/(6) - (1)/(2) = -(1)/(3)`

and

`x_2=(1)/(6) + (1)/(2) = (2)/(3)`