Completing the square part three

So now we can complete the square to solve quadratic equations with positive and negative coefficients for the x term. But what if the coefficient for the x^2 term is greater than 1?

The standard form of a quadratic equation is ax^2 + bx + c = 0 where a, b and c are positive or negative constants.

We have learnt to solve equations where a = 1 and b is positive or negative but what if a is greater than 1? We haven’t tried solving this situation yet.

Let’s try solving 2x^2 + 8x + 4 = 0

This is easy. You just divide everything by two and then proceed as normal.

x^2 + 4x + 2 = 0

(x + 2)^2 -4 + 2 = 0

(x + 2)^2 - 2 = 0

(x + 2)^2 = 2

sqrt{(x + 2)^2} = pmsqrt{2}

x + 2 = pmsqrt{2}

x = -2pmsqrt{2}