Solve quadratic equations · lesson

Taking square roots

💡 If x² equals a positive number, x is plus or minus its square root, so there are two solutions.

If the equation is x² = k, then x = √k or x = −√k. Both square to k.

Get x² alone first. For (x + h)² = k, take the root of both sides, then solve both cases.

🧠 Let's work it out

Solve (x + 1)² = 16.

Take the square root of both sides: x + 1 = 4 or x + 1 = −4.
Subtract 1: x = 3 or x = −5.
Another example: 3x² − 27 = 0 gives x² = 9, so x = 3 or x = −3.

What do you think comes next?

quadratic equation: an equation with an x² term and no higher power
±: plus or minus: both the positive and negative value

🔗 Khan Academy: Solving quadratic equations · Khan Academy ↗

Solving by factoring

💡 Get zero on one side, factor, and set each factor equal to zero.

Move everything to one side so the other side is 0. Then factor.

Each factor set to 0 gives one solution.

If x is a common factor, factor it out. Never divide both sides by x: you'd lose the solution x = 0.

If x² has a number in front, factor it like (2x − 1)(x + 3) = 0. Then 2x − 1 = 0 gives x = 0.5, and x + 3 = 0 gives x = −3. Solutions can be decimals.

🧠 Let's work it out

Solve x² + 5x = 0.

Factor out x: x(x + 5) = 0.
Set each factor to 0: x = 0 or x + 5 = 0.
So x = 0 or x = −5.
Another example: x² − 7x + 10 = 0 factors as (x − 2)(x − 5), so x = 2 or 5.

What do you think comes next?

zero product property: if a product equals 0, one of the factors must be 0

Just one solution

💡 A perfect square trinomial set to zero has only one solution, because both factors are the same.

If the quadratic factors as (x − r)², both factors give the same x. So there is only one solution.

On a graph, the parabola just touches the x-axis at that point.

🧠 Let's work it out

Solve x² + 10x + 25 = 0.

It's a perfect square: (x + 5)² = 0.
So x + 5 = 0, and x = −5.

What do you think comes next?

double root: a solution that comes from a repeated factor

The quadratic formula

💡 For ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) ÷ 2a works every time.

When a quadratic won't factor nicely, use the formula. First read a, b, and c from ax² + bx + c = 0.

Find b² − 4ac first, then its square root. Then compute both the + and − versions.

Round each answer as the question asks.

🧠 Let's work it out

Solve x² − 4x + 2 = 0.

Read the coefficients: a = 1, b = −4, c = 2.
Find b² − 4ac = 16 − 8 = 8.
Use the formula: x = (4 ± √8) ÷ 2 = 2 ± √2.
√2 ≈ 1.414, so x ≈ 3.41 or 0.59.

What do you think comes next?

quadratic formula: x = (−b ± √(b² − 4ac)) ÷ 2a
discriminant: b² − 4ac, the part under the square root