Question

How do you solve the quadratic equation by completing the square: `3x^2-9x=-1`?

Answer 1

`x_(1,2) = 3/2 +- sqrt(69)/6`

Explanation:

Starting from the general form of the quadratic equation

`ax^2 + bx + c = 0`

divide verything by `a` to get it to the form

`x^2 + b/ax = -c/a`

In your cae, `a=3`, which means that you get

`(color(red)(cancel(color(black)(3))) x^2)/color(red)(cancel(color(black)(3))) - 9/3x = -1/3`

`x^2 - 3x = -1/3`

To solve this quadratic by completing the square, you need to write the left side of the equation as the square of a binomial. This can be done by adding a term to both sides of the equation.

The coefficient of the `x`-term will help you determine what you need to add to both sides.

So, divide this coefficient by 2, then square the result to get

`(-3)/2 = -3/2`, then

`(-3/2)^2 = 9/4`

Adding `9/4` to both sides of the equation will produce

`x^2 - 3x + 9/4 = -1/3 + 9/4 = 23/12`

The general formula for the square of a binomial looks like this

`color(blue)((x + n)^2 = x^2 + 2n + n^2)`

You can use this formula to rewrite the left side of the equation as

`x^2 - 3x + 9/4 = x^2 + 2 * (-3/2)x + (-3/2)^2`

`x^2 - 3x + 9/4 = (x-3/2)^2`

Your quadratic becomes

`(x-3/2)^2 = 23/12`

Take the square root of both sides to get

`sqrt((x-3/2)^2) = sqrt(23/12)`

`x-3/2 = +-sqrt(23)/sqrt(12) = +- (sqrt(23) * sqrt(3))/(2 * 3) = +- sqrt(69)/6`

The two solutions to your quadratic equation will thus be

`x_1 = color(green)(3/2 + sqrt(69)/6)` and `x_2 = color(green)(3/2 - sqrt(69)/6)`