Question

How do you solve `2x ^ { 2} - 56x + 6= 0` by completing the square?

Answer 1

`color(blue)(x = 14 + sqrt(193), x = 14 - sqrt(193))`

Explanation:

We have the following quadratic equation in our problem:

`color(red)(2x^2 - 56x + 6 = 0)` `color(blue)(...Equation.1)`

Note that the coefficient of `x^2 term` is greater than 1

`color(green)(Step.1)`

In our `color(blue)(...Equation.1)`, we are going move the constant term from the left hand side(LHS) to the right side(RHS).

The constant term is `+6` in our `color(blue)(...Equation.1)`

Add `color(blue){(-6)}` to both sides of our equation.

`color(red)(2x^2 - 56x + 6 - 6 = 0-6)`

`color(red)(2x^2 - 56x = -6)` `color(blue)(...Equation.2)`

`color(green)(Step.2)`

Since the coefficient of `(x^2 term)` is greater than 1, divide out every term of our equation by `2`. This process will make it easier to complete the square

So, our `color(blue)(...Equation.2)` will now become:

`color(red)((2x^2)/2 - (56x)/2 = -6/2)`

Simplifying we get,

`color(red)(x^2 - 28x = -3)` `color(blue)(...Equation.3)`

`color(green)(Step.3)`

We are going to add a term to both sides of equation as follows:

`color(red)(x^2 - 28x + square = -3 + square)`

What are we going to write in the box?

Divide the coefficient of `(-28x)` by `2` and square it.

`((-28)/2)^2`

We get, `14^2 = 196`

This value of `196` goes into the box.

Hence, we get

`color(red)(x^2 - 28x + 196 = -3 + 196)`

`rArr color(red)(x^2 - 28x + 196 = 193)` `color(blue)(...Equation.4)`

`color(green)(Step.4)`

We can now write the LHS in `color(blue)(...Equation.4)` as a Perfect Square

Divide the coefficient of the term `-28x` by 2 and use it to write LHS as a perfect square as shown below:

`rArr color(red)((x - 14)^2 = 193)` `color(blue)(...Equation.5)`

`color(green)(Step.5)`

Take the square root on both sides to simplify.

`color(blue)(...Equation.5)` will now become

`color(red)(sqrt((x-14)^2) = +-sqrt(193))`

We notice that on the LHS both the square root and the square will cancel out to yield

`color(red)((x-14) = +-sqrt(193))` `color(blue)(...Equation.6)`

`color(green)(Step.6)`

Using our `color(blue)(Equation.6)` we get two solutions

`color(red)((x-14) = +sqrt(193) and (x-14) = -sqrt(193)`

Hence, rearranging the terms, our final solutions are

`color(blue)(x = 14 + sqrt(193), x = 14 - sqrt(193)))`

I hope this helps.