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Lesson overview

Algebra - solving quadratic equations

The quadratic formula is a powerful tool that solves any quadratic equation of the form $ax^2 + bx + c = 0$ . The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ The expression $b^2 - 4ac$ is called the discriminant. It tells us how many solutions the equation has: if the discriminant is positive, there are two different real solutions; if it is zero, there is one repeated root; if it is negative, there are no real solutions. The quadratic formula works for all quadratics, including those that don't factorise easily.

Using the quadratic formula

Solve $x^2 + 5x + 6 = 0$ using the quadratic formula.

Identify $a$ , $b$ , and $c$ carefully. Remember the $\pm$ symbol means you get two solutions.

When the quadratic doesn't factorise

Solve $2x^2 - 3x - 1 = 0$ using the quadratic formula.

When $b$ is negative, be careful with signs. The discriminant $b^2 - 4ac = 9 - 4(2)(-1) = 9 + 8 = 17$ .

Understanding the discriminant

Find the discriminant of $x^2 + 2x + 5 = 0$ and state how many real solutions exist.

A negative discriminant means the parabola doesn't cross the x-axis.

Repeated roots (equal solutions)

Solve $x^2 + 6x + 9 = 0$ and interpret the discriminant.

When the discriminant is 0, the parabola touches the x-axis at exactly one point.

Worked example

Solve $3x^2 + 2x - 1 = 0$ using the quadratic formula. Give your answers as exact values (surds if needed) and also as decimals to 2 decimal places.

Identify the coefficients: $a = 3$ , $b = 2$ , $c = -1$ .
Calculate the discriminant: $b^2 - 4ac = 4 - 4(3)(-1) = 4 + 12 = 16$ .
Since the discriminant is positive (16 > 0), there are two different real solutions.
Substitute into the formula: $x = \frac{-2 \pm \sqrt{16}}{2(3)} = \frac{-2 \pm 4}{6}$ .
Using the + sign: $x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$ (or 0.33 to 2 d.p.).
Using the - sign: $x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$ .
The solutions are $x = \frac{1}{3}$ and $x = -1$ .

Try it

For each quadratic, identify $a$, $b$, and $c$, then use the formula. Check your discriminant calculation.

Question 1

Solve $x^2 - 5x + 6 = 0$ using the quadratic formula.

💡 Identify $a = 1$ , $b = -5$ , $c = 6$ . Calculate the discriminant: $25 - 24 = 1$ .

Question 2

What is the discriminant of $x^2 + 4x + 4 = 0$ ? How many real solutions does this equation have?

💡 Calculate $b^2 - 4ac = 16 - 16 = 0$ . When the discriminant is 0, there is one repeated root.

Question 3

Solve $2x^2 + x - 3 = 0$ using the quadratic formula.

💡 Here $a = 2$ , $b = 1$ , $c = -3$ . Discriminant = $1 - 4(2)(-3) = 1 + 24 = 25$ . So $x = \frac{-1 \pm 5}{4}$ .

Question 4

The equation $x^2 + 2x + 5 = 0$ has which of the following?

💡 Calculate the discriminant: $4 - 4(1)(5) = 4 - 20 = -16$ . A negative discriminant means no real solutions.

Question 5

Solve $3x^2 - 6x = 0$ using the quadratic formula.

💡 First rewrite as $3x^2 - 6x + 0 = 0$ . Then $a = 3$ , $b = -6$ , $c = 0$ . Discriminant = $36 - 0 = 36$ .

Common mistakes

Watch for these when working through the lesson.

Misidentifying the values of $a$ , $b$ , and $c$ . Be careful with signs, especially when $b$ or $c$ is negative.
Calculating the discriminant incorrectly. Remember it is $b^2 - 4ac$ , not $b^2 + 4ac$ .
Forgetting the $\pm$ in the formula, which leads to finding only one solution instead of two.
Not simplifying the final answer fully. For example, $\frac{6}{4}$ should be simplified to $\frac{3}{2}$ .

The quadratic formula

Lesson overview

Worked example

Try it

Common mistakes

Related topics

Independent practice will plug in here