Solve quadratic equations using the quadratic formula and by completing the square, and express a quadratic in completed-square form

What this dot point is asking

SEAB wants you to solve quadratic equations that do not factorise neatly, using the quadratic formula and the method of completing the square, and to express a quadratic in completed-square form to read off its minimum or maximum. These methods handle every quadratic, even when the roots are irrational.

The answer

The quadratic formula

For any quadratic $ax^2 + bx + c = 0$, the solutions are:

Identify $a$, $b$ and $c$ carefully, including signs, substitute, and evaluate the two values from the plus and the minus. Give the answer to the accuracy the question asks, often two or three significant figures or two decimal places.

The discriminant

The quantity under the root, $b^2 - 4ac$, is the discriminant. If it is positive there are two real solutions, if it is zero there is one repeated solution, and if it is negative there are no real solutions because you cannot take the square root of a negative number at this level.

Completing the square to solve

Completing the square rewrites $x^2 + bx$ as $\left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2$. Once the equation is a perfect square equal to a number, take the square root of both sides, remembering the plus and minus, then solve.

Completing the square for the minimum

Written as $(x + a)^2 + b$, a quadratic has its minimum value $b$ when the squared term is zero, at $x = -a$. This is how completing the square locates the turning point of a parabola without calculus, useful in the graphs strand.

Examples in context

Example 1. A projectile's flight time. The height of a ball might be modelled by a quadratic in time, $h = -5t^2 + 20t + 1$. Setting $h = 0$ and using the formula finds when it lands, a calculation where the roots are rarely whole numbers, so the formula is the natural tool.

Example 2. Minimum cost. If a cost function is $C = x^2 - 8x + 30$, completing the square gives $(x - 4)^2 + 14$, showing the least cost is $14$ when $x = 4$. Completing the square reads off the optimum directly.

Try this

Q1. Use the formula to solve $x^2 + 3x - 2 = 0$, giving answers to 2 decimal places. [3 marks]

• Cue. $x = \dfrac{-3 \pm \sqrt{9 + 8}}{2} = \dfrac{-3 \pm \sqrt{17}}{2}$, giving $0.56$ or $-3.56$.

Q2. Express $x^2 - 4x + 7$ in the form $(x - a)^2 + b$. [2 marks]

• Cue. Half of $4$ is $2$: $(x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$.

Q3. State the number of real solutions of $x^2 + 2x + 5 = 0$ and justify. [2 marks]

• Cue. Discriminant $= 2^2 - 4(1)(5) = 4 - 20 = -16 < 0$, so there are no real solutions.