We are going to use the completing the square method to help us prove this.
    ax2 + bx + c = 0

    x2 + (b/a)x + c/a = 0

    (x2 + (b/a)x + (b/2a)2) - (b/2a)2 + c/a = 0

    (x + b/2a)2 - b2/4a2 + c/a = 0

    (x + b/2a)2 = b2/4a2 - c/a

    (x + b/2a)2 = (b2 - 4ac)/4a2

Now 4a2 > 0, so if b2 - 4ac < 0 this implies that (b2 - 4ac)/4a2 < 0. This means there is no real number x which solves the equation. Hence (i) holds true.

If b2 - 4ac = 0, then (x + b/2a)2 = 0 So x+ b/2a = 0 implies that x = -b/2a. Therefore (ii) is true.

If b2 - 4ac > 0, then (b2 - 4ac)/4a2 > 0 and there are two real numbers SQRT(b^2 - 4ac)/2a and -SQRT(b^2 - 4ac)/2a whose squares are equal to (b2 - 4ac)/4a2

Hence x + b/2a = SQRT(b^2 - 4ac)/2a and x + b/2a = -SQRT(b^2 - 4ac)/2a
This implies that x = (-b + SQRT(b^2 - 4ac))/2a and (-b - SQRT(b^2 - 4ac))/2a which proves (iii).

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