Question

Can anyone teach me how to solve a Quadratic equation by completing the square? Thanks..

Answer 1

The idea behind completing the square is to add or subtract a constant to obtain the form `(x-h)^2` and then take a square root to be left with a linear equation. Let's do a concrete example first.

Starting from `2x^2-7x-4=0`

Step 1: Divide both sides by `2` to obtain `x^2` as the first term

`x^2-7/2x-2 = 0`

Step 2: Add `2` to both sides to isolate the `x` terms.

`x^2-7/2x = 2`

Step 3: Add a constant to both sides which will allow us to factor the left hand side as `(x-h)^2`. Noting that `(x-h)^2 = x^2-2h+h^2` we have `-2h = -7/2` and thus `h = 7/4`, meaning we add `(7/4)^2 = 49/16` to both sides.

`x^2-7/2+49/16 = 81/16`

Step 4: Factor the left hand side

`(x-7/4)^2 = 81/16`

Step 5: Take the square root of both sides. Remember to account for both positive and negative roots.

`x-7/4 = +-sqrt(81/16) = +-9/4`

Step 6: Solve the remaining linear equation:

`x = 7/4 +- 9/4 = 1/4(7+-9)`

`=> x = 4` or `x = -1/2`

The real trick here is observing in step 3 that the constant we need to add is equal to the square of half of the coefficient of `x`.

---

Let's see what happens if we apply this to a general quadratic equation.

`ax^2 + bx + c = 0`

`=> x^2 + b/ax + c/a = 0`

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

`=> x^2 + b/ax + (b/(2a))^2 = -c/a + (b/(2a))^2 = b^2/(4a^2)-c/a`

`=> (x+b/(2a))^2 = (b^2-4ac)/(4a^2)`

`=> x+b/(2a) = +-sqrt((b^2-4ac)/(4a^2)) = +-sqrt(b^2-4ac)/(2a)`

`=> x = -b/(2a) +- sqrt(b^2-4ac)/(2a)`

`=(-b+-sqrt(b^2-4ac))/(2a)`

And we have just derived the quadratic formula.