Question

How do you solve by completing the square: `x^2 – 4x – 60 = 0`?

Answer 1

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
`(x+a)^2=b`
So, if `b\ge 0`, you can take the square root at both sides to get
`x+a=\pm\sqrt{b}`
and conclude `x=\pm\sqrt{b}-a`.

Now, we have `(x+a)^2=x^2+2ax+a^2`. Since you equation starts with `x^2-4x`, this means that `2ax=-4x`, and so `a=-2`.
Adding `64` at both sides, we have
`x^2-4x+4=64`
Which is the form we wanted, because now we have
`(x-2)^2=64`
Which leads us to
`x-2=\pm\sqrt{64}=\pm 8` and finally `x=\pm8+2`, which means that the two solutions are `-8+2=-6` and `8+2=10`

Answer 2

• First, we Transpose the Constant to one side of the equation.
  Transposing `-60` to the other side we get:
  `x^2-4x = 60`

• Application of `(a-b)^2 = a^2 - 2ab + b^2`
  We look at the Co-efficient of `x`. It's `-4`
  We take half of this number (including the sign), giving us `–2`
  We square this value to get `(-2)^2 = 4`. We add this number to BOTH sides of the Equation.
  `x^2-4x+4 = 60+4`
  `x^2-4x+4 = 64`
  The Left Hand side `x^2-4x+4` is in the form `a^2 - 2ab + b^2`
  where `a` is `x`, and `b` is `2`

• The equation can be written as
  `(x-2)^2 = 64`

So `(x-2)` can take either `8` or `-8` as a value. That's because squaring both will give us 64.

`x-2 = 8` (or) `x-2 = -8`
`x = 10` (or) `x = -6`

• Solution : `x = 10,-6`

• Verify your answer by substituting these values in the Original Equation `x^2- 4x - 60 = 0`
  The Left hand Side is `x^2- 4x - 60`, and the Right Hand Side is `0`

If x = 10,
Left Hand Side
`= (10)^2 - 4(10) - 60`
`= 100 - 40 - 60`
`= 0` (Is Equal to the Right Hand Side)

If x = -6,
Left Hand Side
`= (-6)^2 - 4(-6) - 60`
`= 36 + 24 - 60`
`= 0` (Is Equal to the Right Hand Side)

Both the solutions are verified. Our solution `x = 10,-6` is correct.