The Quadratic Formula

Grade 6

The formula that always works

When a quadratic will not factorise cleanly, the quadratic formula solves it every time. For any equation in the form ax² + bx + c = 0:

x = (−b ± √(b² − 4ac)) / (2a)

The ± means there are usually two solutions. Write both out clearly.

Identifying a, b, and c

Before using the formula, write the equation in standard form ax² + bx + c = 0. Then identify a (coefficient of x²), b (coefficient of x), and c (the constant). Get the signs right — if your equation is 3x² − 2x − 5 = 0, then a = 3, b = −2, c = −5.

Worked example

EXAMPLE

Solve x² − 4x − 2 = 0. Give your answers to 2 decimal places.

a = 1, b = −4, c = −2. Calculate discriminant: b² − 4ac = (−4)² − 4(1)(−2) = 16 + 8 = 24. √24 ≈ 4.899. x = (4 ± 4.899) / 2. So x = (4 + 4.899) / 2 ≈ 4.45 or x = (4 − 4.899) / 2 ≈ −0.45. x ≈ 4.45 or x ≈ −0.45 (to 2 d.p.).

The discriminant tells you how many solutions

The discriminant is b² − 4ac. If it is positive: two solutions. If zero: one repeated solution. If negative: no real solutions (the parabola does not cross the x-axis).

Check the discriminant before applying the formula. If it is negative, you know immediately there are no real answers — no need to continue.

Try this one

A right-angled triangle has legs of length \( x \) cm and \( (x + 2) \) cm, and a hypotenuse of \( 10 \) cm.

(a) Show that \( x^2 + 2x - 48 = 0 \). [3]

(b) Find the length of the shorter leg. [2]

Nova's hint:

This combines Pythagoras' theorem with solving a quadratic.

Ask yourself: 'how do the squares of the two legs relate to the hypotenuse?' Expand \( (x+2)^2 \) carefully, then collect terms.

Try to Solve it with Nova?