- Maths
- Classes 9-10
- 5 min read
Quadratic equations
Solve ax² + bx + c = 0 by factorising and by the quadratic formula, and use the discriminant to predict how many real roots there are.
Let us begin
The big idea
A quadratic equation has an x² term, and its graph is a U-shaped curve called a parabola. Solving the equation means finding where that curve crosses the x-axis. There can be two crossings, one touch, or none at all.
A quadratic equation is ax² + bx + c = 0 with a not equal to 0. Its solutions are called roots. Because the highest power is 2, there are at most two roots.
Method 1, factorising: write the left side as a product of two brackets. If (x − 2)(x − 3) = 0 then one bracket must be zero, so x = 2 or x = 3.
Method 2, the quadratic formula, which always works: x = (−b ± √(b² − 4ac)) / 2a. The part under the root, D = b² − 4ac, is called the discriminant.
The discriminant tells you the nature of the roots before you solve. If D is positive there are two different real roots. If D = 0 the two roots are equal (the parabola just touches the axis). If D is negative there are no real roots (the parabola never reaches the axis).
Worked examples
See how it works
By factorising: x² − 5x + 6 = 0
Two numbers with product 6 and sum −5 are −2 and −3. So (x − 2)(x − 3) = 0 and the roots are x = 2 and x = 3. Check: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.
By factorising with a leading coefficient: 2x² + x − 6 = 0
Product a × c = −12, sum 1: the numbers are 4 and −3. So 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2) = 0. The roots are x = 3/2 and x = −2.
By the formula: x² + 4x + 1 = 0
Here a = 1, b = 4, c = 1, so D = 16 − 4 = 12. Then x = (−4 ± √12) / 2 = (−4 ± 2√3) / 2 = −2 ± √3. The roots are about −0.27 and −3.73.
A word problem
The product of two consecutive positive whole numbers is 132. Call them n and n + 1: n(n + 1) = 132, so n² + n − 132 = 0. Numbers with product −132 and sum 1 are 12 and −11, so (n + 12)(n − 11) = 0. Since n is positive, n = 11 and the numbers are 11 and 12.
Try it at home
Plot a parabola by hand
You will need
- Squared or ruled paper
- A pencil and ruler
What to do
- Make a table for y = x² − 5x + 6 with x = 0, 1, 2, 2.5, 3, 4 and 5. You should get y = 6, 2, 0, −0.25, 0, 2, 6.
- Draw axes with x from 0 to 5 and y from −1 to 6, and plot the seven points.
- Join them with a smooth U-shaped curve (do not use the ruler for this part).
- Read off where the curve crosses the x-axis. Those two x values are the roots of x² − 5x + 6 = 0.
- Now try y = x² − 4x + 4 with the same x values. Notice the curve only touches the axis once, at x = 2, because D = 16 − 16 = 0.
Practice sheet
Check what you learned
Practice sheet: Quadratic equations (Classes 9-10, Maths)
- Is x² − 4 = 0 a quadratic equation? What are its roots?
- Solve x² − 7x + 12 = 0 by factorising.
- Solve x² + 2x − 8 = 0.
- Find the discriminant of 2x² − 3x + 1 = 0 and then its roots.
- What is the nature of the roots of x² + x + 1 = 0?
- Solve x² − 6x + 9 = 0. What do you notice about the roots?
- The square of a number minus three times the number is 10. Find the positive number.
- Solve x² = 49.
- Solve 3x² − 12 = 0.
- Write the quadratic formula.
Answer key
Answers: Quadratic equations
- Yes, degree 2. Roots: x = 2 and x = −2.
- (x − 3)(x − 4) = 0, so x = 3 or x = 4.
- (x + 4)(x − 2) = 0, so x = −4 or x = 2.
- D = 9 − 8 = 1. Roots: x = (3 ± 1)/4, so x = 1 or x = 1/2.
- D = 1 − 4 = −3, which is negative, so there are no real roots.
- (x − 3)² = 0, so x = 3. The two roots are equal because D = 36 − 36 = 0.
- x² − 3x − 10 = 0, so (x − 5)(x + 2) = 0. The positive number is 5.
- x = 7 or x = −7.
- x² = 4, so x = 2 or x = −2.
- x = (−b ± √(b² − 4ac)) / 2a.