Completing the square and quadratic formula practice | Year 10

Choose a method that suits the coefficients. Then use the discriminant as a check on the number of real roots, keeping exact surds until the question asks for decimals.

Complete a square by balancing both sides of an equation.
Substitute signed coefficients into the quadratic formula.
Interpret positive, zero and negative discriminants over the real numbers.

The idea

Completing the square rewrites a quadratic using a squared bracket. For x² + bx, add (b/2)² to make (x + b/2)², balancing the other side as well. The quadratic formula works for every quadratic with a nonzero squared coefficient. Its discriminant b² − 4ac tells us how many distinct real roots to expect.

Before you begin: Expand a squared bracket, manipulate equations and simplify square roots.

Worked examples

01 Start with the method

Solve exactly by completing the square.

x^{2} + 4 x - 1 = 0

Step 1

Move the constant to the other side. Half the coefficient of x is 2, and its square is 4.

x^{2} + 4 x = 1,

{(\frac{4}{2})}^{2} = 4

Step 2

Add 4 to both sides, then rewrite the left side as a square.

x^{2} + 4 x + 4 = 5

\Rightarrow {(x + 2)}^{2} = 5

Step 3

A positive square has two possible square roots. Keep both signs.

x + 2 = \pm \sqrt{5}

\Rightarrow x = - 2 \pm \sqrt{5}

Step 4

The two exact roots lie on opposite sides of −2. Expanding the squared form recovers the original equation.

{(x + 2)}^{2} - 5 = x^{2} + 4 x - 1

A mistake to avoid

Adding a term on only one side changes the equation. Also, taking the square root requires ± when both real roots are being found.

02Use the formula for a non-monic quadratic

Solve exactly using the quadratic formula.

2 x^{2} - 3 x - 1 = 0

Step 1

Identify the signed coefficients. The denominator of the formula is 2a, not a.

a = 2

b = - 3

c = - 1,

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Step 2

Evaluate the discriminant with brackets around negative coefficients.

b^{2} - 4 a c = {(- 3)}^{2} - 4 (2) (- 1) = 17

Step 3

Substitute. Seventeen is not a perfect square, so keep its square root exact.

x = \frac{3 \pm \sqrt{17}}{4}

Result

x = \frac{3 + \sqrt{17}}{4} or x = \frac{3 - \sqrt{17}}{4}

03Check whether real roots exist

How many real solutions does this equation have?

x^{2} + 2 x + 5 = 0

Step 1

Compute the discriminant before attempting a real square root.

Δ = 2^{2} - 4 (1) (5) = - 16

Step 2

The discriminant is negative, so the formula cannot give real roots. Completing the square explains why.

{(x + 1)}^{2} + 4 = 0

\Rightarrow {(x + 1)}^{2} = - 4

Step 3

A real number’s square cannot be negative. The conclusion is about real solutions; complex solutions are a separate extension.

{(x + 1)}^{2} \geq 0 for real x

Result

No real solutions

Practice questions with answers

Work these out first, then open each solution to compare your method. These questions are untimed.

Question 1

Complete the square and give both exact roots.

x^{2} - 6 x + 2 = 0

Show solution 1

{(x - 3)}^{2} = 7

\Rightarrow x = 3 \pm \sqrt{7}

Move 2 to the right, then add 9 to both sides. Half of −6 is −3, so the squared bracket is (x − 3)².

Question 2

Use the formula and simplify the exact roots.

3 x^{2} + 2 x - 2 = 0

Show solution 2

x = \frac{- 2 \pm \sqrt{28}}{6} = \frac{- 1 \pm \sqrt{7}}{3}

The discriminant is 4 + 24 = 28. Since √28 = 2√7, divide the whole numerator and denominator by 2.

Question 3

Find the root and describe the discriminant.

x^{2} - 8 x + 16 = 0

Show solution 3

{(x - 4)}^{2} = 0,

x = 4,

Δ = 0

A zero discriminant gives one distinct repeated real root. Both signs in the formula lead to 4.

Question 4

Determine the number of real roots.

x^{2} + 4 x + 8 = 0

Show solution 4

Δ = 16 - 32 = - 16

\Rightarrow no real roots

A negative discriminant means there is no real value for the square root in the formula.

Question 5

Complete the square and solve exactly.

x^{2} - 2 x - 4 = 0

Show solution 5

{(x - 1)}^{2} = 5

\Rightarrow x = 1 \pm \sqrt{5}

Move −4 to the right, then add 1 to both sides. The resulting squared bracket is x − 1.

Common questions

When is completing the square convenient?

It is particularly direct when the squared coefficient is one and the x coefficient is even. The method still works with other coefficients, but may involve fractions.

What does the discriminant tell me?

For real coefficients, a positive discriminant gives two distinct real roots, zero gives a repeated real root, and a negative discriminant gives no real roots.

Should I turn the square root into a decimal?

Keep the surd when an exact answer is requested. If a decimal approximation is requested, round only at the end to the stated accuracy.