Question

How do you solve `3x^2-2x-3=0` by completing the square?

Answer 1

`x = +-sqrt(10/9) + 1/3`, or `~~ 1.39` and `~~-0.72`

Explanation:

First things first, in order to complete the square, the leading coefficient (`color(red)(3)x^2-2x-3`) must be `1`. To do that, we need to factor out a `3` from the equation.

`3x^2-2x-3 =0`

`3(x^2-2/3x-1)=0`

Now we have the beginning. Completing the square can look scary, but it's really just a process, and if you understand the steps, it becomes pretty simple.

The first step is getting the leading coefficient to `1`. After that, we need to take the middle term, `-2/3` and "do some stuff with it" (you'll see in a minute)

So, we'll take `(-2/3)` and divide it by `2`, which gives us `-1/3`. Now we square the solution, which equals `1/9`.

We did all of this because we needed to find the value that will make the left side of our equation, `x^2-2/3x-1 =0`, a perfect square, which is `1/9`.

Now that we have our missing value, we need to add it to our equation.

`x^2-2/3x-1 + color(red)(1/9) " "=0`

But wait!! We can't just add a random number into an equation! An equation is all about balance (the root word is equal). You can't just introduce a new value. But.... if you add `500`, and then immediately subtract `500`, the final result is `0`.

So, if we add `color(red)(1/9)`, and then subtract `color(red)(1/9)`, then technically we haven't changed anything

`x^2-2/3x-1 + color(red)(1/9) + color(red)(-1/9)=0`

Let's re-order this:

`color(green)(x^2-2/3x+1/9) color(blue)(-1 -1/9)=0`

`color(green)(x^2-2/3x+1/9)` is a perfect square (that was the whole point of all of this, after all). Let's factorise it.

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

Let's simplify this: `color(blue)(-1 -1/9)` equals `color(blue)(-10/9)`

So, now we have `color(green)((x-1/3)^2)color(blue)(-10/9)=0`

`* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *`

Now, let's solve this puppy!

`(x-1/3)^2-10/9=0`

add `10/9` to both sides

`(x-1/3)^2 = 10/9`

take a square root of both sides

`sqrt((x-1/3)^2) = +-sqrt(10/9)`

`x-1/3 = +-sqrt(10/9)`

add `1/3` to both sides

`x = +-sqrt(10/9) + 1/3`,

or `~~ 1.39` and `~~-0.72`

Answer 2

`x=(sqrt10+1)/3`
`x=(-(sqrt10)+1)/3`

Explanation:

Given -

`3x^2-2x-3=0`

Take the constant to the right

`3x^2-2x=3`

Divide both sides by the coefficient of `x^2`

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

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

Take half of the coefficient of `x` and square it

Half the coefficient of `x=-2/3-:2=-2/3xx1/2=-2/6`

Square of `-2/6=4/36=1/9`

Add `1/9` to both the sides

`x^2-2/3x+1/9=1+1/9=(9+1)/9=10/9`

`(x-1/3)^2=10/9`

Taking square root on both sides

`x-1/3=+-sqrt (10/9)=+-sqrt10/3`
`x=+-sqrt10/3+1/3= +-(sqrt10+1)/3`

`x=(sqrt10+1)/3`
`x=(-(sqrt10)+1)/3`