Grade by Grade
A. Characteristics of Functions · Quadratic Functions

Exact Zeros: Solving Quadratic Equations with Radicals

A2.1About 70 minutes
Learning goals
  • I can find exact zeros of a quadratic function by the square root method, completing the square or the quadratic formula.
  • I can simplify answers from the quadratic formula fully.
  • I can check exact roots with their sum and product.
  • I can decide which roots make sense in a real-world problem.

Words to know

root
A solution of an equation. The roots of f(x) = 0 are the zeros of f.
quadratic formula
x = (−b ± √(b² − 4ac))/(2a), the solutions of ax² + bx + c = 0.
exact value
A value written without rounding, such as 1 + √5.
approximate value
A rounded decimal, such as 3.24.

When a quadratic will not factor

The curriculum gives this example: a quadratic equation has roots 1 + √5 and 1 − √5. Where do numbers like that come from? They are the zeros of f(x) = x² − 2x − 4, which does not factor over the integers. No two integers multiply to −4 and add to −2.

In Grade 10 you solved such equations with the quadratic formula and usually rounded the answers. In Grade 11 you will give exact answers, in simplest radical form, using the skills from the last lesson. Exact zeros let you write exact equations, check answers without rounding error, and see patterns that decimals hide.

The quadratic formula (from Grade 10)

The solutions of ax² + bx + c = 0, with a ≠ 0, are

x = (−b ± √(b² − 4ac)) / (2a).

The expression under the root, b² − 4ac, is the discriminant. The next lesson uses it to count zeros.

Three ways to get exact zeros

  1. Square root method. If the function is in vertex form, f(x) = a(x − h)² + k, set it equal to 0, isolate the square and take ± square roots.
  2. Completing the square. Turn standard form into vertex form, then use method 1.
  3. Quadratic formula. Works on every quadratic equation in standard form.

In every case, finish by simplifying the radical and dividing out common factors.

Worked example

Example 1: Square root method

Find the exact zeros of f(x) = 2(x − 1)² − 6.

  1. Set f(x) = 0: 2(x − 1)² − 6 = 0.
  2. Isolate the square: 2(x − 1)² = 6, so (x − 1)² = 3.
  3. Take both square roots: x − 1 = ±√3.
  4. x = 1 ± √3. Decimal check: x ≈ 2.73 or x ≈ −0.73.

x = 1 + √3 or x = 1 − √3.

Worked example

Example 2: The quadratic formula

Solve x² − 2x − 4 = 0 exactly.

  1. a = 1, b = −2, c = −4.
  2. Discriminant: b² − 4ac = (−2)² − 4(1)(−4) = 4 + 16 = 20.
  3. x = (2 ± √20) / 2.
  4. Simplify the radical: √20 = √4 × √5 = 2√5, so x = (2 ± 2√5) / 2.
  5. Divide out the common factor 2: x = 1 ± √5.

x = 1 + √5 ≈ 3.24 or x = 1 − √5 ≈ −1.24.

Simplifying the roots of x² − 2x − 4 = 0(2 ± √20)/2formula result(2 ± 2√5)/2√20 = √4 × √52(1 ± √5)/2common factor 21 ± √5simplest form
Finishing an answer from the quadratic formula: simplify the radical, then divide every term of the numerator by the common factor.
−3−2−1012345−6−4−20246xy1 − √5 ≈ −1.241 + √5 ≈ 3.24(1, −5)
The graph of f(x) = x² − 2x − 4. Its exact zeros 1 − √5 and 1 + √5 sit √5 ≈ 2.24 units on either side of the axis x = 1.
Worked example

Example 3: Completing the square

Solve x² + 6x + 4 = 0 by completing the square.

  1. Move the constant: x² + 6x = −4.
  2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 5.
  3. (x + 3)² = 5, so x + 3 = ±√5.
  4. x = −3 ± √5 (about −0.76 and −5.24).

x = −3 + √5 or x = −3 − √5.

Worked example

Example 4: a ≠ 1

Solve 2x² + 6x + 1 = 0 exactly.

  1. a = 2, b = 6, c = 1. Discriminant: 36 − 4(2)(1) = 36 − 8 = 28.
  2. x = (−6 ± √28) / 4.
  3. √28 = √4 × √7 = 2√7, so x = (−6 ± 2√7) / 4.
  4. Every term has a factor of 2: x = 2(−3 ± √7) / 4 = (−3 ± √7) / 2.
  5. Decimals: (−3 + 2.646)/2 ≈ −0.18 and (−3 − 2.646)/2 ≈ −2.82.

x = (−3 + √7)/2 or x = (−3 − √7)/2.

Worked example

Example 5: Rearrange first

Solve 3x² = 4x + 2 exactly.

  1. Write in standard form: 3x² − 4x − 2 = 0. So a = 3, b = −4, c = −2.
  2. Discriminant: 16 − 4(3)(−2) = 16 + 24 = 40.
  3. x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.
  4. Decimals: about 1.72 and −0.39.

x = (2 ± √10)/3.

A quick check for exact roots

For ax² + bx + c = 0, the two roots always add to −b/a and multiply to c/a. For x² − 2x − 4 = 0: (1 + √5) + (1 − √5) = 2 = −(−2)/1 ✓ and (1 + √5)(1 − √5) = 1 − 5 = −4 = −4/1 ✓. If your roots fail this test, look for an error.

Hands-on activity

The golden rectangle

You need: paper, a ruler, a pencil and scissors (take care with scissors; a grown-up can help) and a calculator.

  1. Solve x² − x − 1 = 0 exactly. You should get x = (1 ± √5)/2. The positive root, (1 + √5)/2 ≈ 1.618, is called the golden ratio.
  2. Cut a rectangle 10 cm wide and 16.2 cm long (16.2 ÷ 10 ≈ 1.618).
  3. Cut a 10 cm by 10 cm square off one end. Measure the leftover rectangle.
  4. Divide its long side by its short side. Compare with 1.618.
  5. Use the calculator to substitute the rounded values x = 1.6, 1.62 and 1.618 into x² − x − 1. Watch the output get closer to 0, but never reach it exactly. Only the exact value (1 + √5)/2 gives 0.

What to look for: the leftover rectangle is 10 cm by 6.2 cm, and 10 ÷ 6.2 ≈ 1.61, almost the same shape as the original. The outputs are about −0.04, 0.004 and −0.0001. That is why exact answers matter.

Just for fun

The golden ratio φ = (1 + √5)/2 has a party trick: to find 1/φ, just subtract 1. Since φ² = φ + 1, dividing by φ gives φ = 1 + 1/φ, so 1/φ = φ − 1 ≈ 0.618. Few numbers are that easy to flip.

Your turn

1

Solve (x − 4)² = 7 exactly.

2

Find the exact zeros of f(x) = 3(x + 2)² − 24.

3

Solve x² + 4x − 1 = 0 by completing the square.

4

Solve x² − 6x + 7 = 0 with the quadratic formula. Simplify fully.

5

Solve 2x² − 2x − 1 = 0 exactly.

6

Find the exact zeros of f(x) = −x² + 8x − 13.

7

Give the zeros from the last question to two decimal places.

More practice: warm-up

1

Solve x² = 18 exactly.

2

Solve (x − 1)² = 9.

3

Find the discriminant of x² + 2x − 2 = 0.

4

Simplify (4 ± √12)/2.

5

Simplify (−6 ± √20)/2.

More practice: on your own

1

Solve 4x² + 4x − 1 = 0 exactly.

2

Solve x² = 10x − 22 exactly.

3

Find the exact zeros of f(x) = 2(x + 3)² − 10.

4

Solve 3x² + 2x − 2 = 0. Give exact and approximate answers.

5

A ball's height is h(t) = −5t² + 10t + 2 metres after t seconds. Find the exact time it lands, then round to the nearest hundredth.

Challenge

1

Verify the sum-and-product check for the roots 1 + √5 and 1 − √5 of x² − 2x − 4 = 0.

2

Solve x² − 2√3x − 1 = 0 exactly.

3

A rectangle's length is 2 m more than its width and its area is 10 m². Find the exact width and length.

Check your understanding

Quiz

1.

What are the exact solutions of x² = 12?

2.

What are the solutions of x² − 4x + 1 = 0?

3.

Which is the correct simplification of (6 ± √8)/2?

4.

What are the zeros of f(x) = (x + 1)² − 6?

5.

Why might you give the zeros as 1 ± √5 instead of 3.24 and −1.24?

6.

Solve x² + 2x − 1 = 0.

For parents and teachers

Big idea: Expectation A2.1 is about the number of zeros of a quadratic function. This lesson supports it by finding the zeros exactly, in simplest radical form, which is also needed for A2.4 (equations from irrational roots).

How to help: Ask your teen to check exact answers with a calculator and with the sum-and-product test.

Common difficulties:

  • Sign errors with −b when b is negative.
  • Forgetting to put b² − 4ac in brackets on the calculator.
  • Dividing only part of the numerator by the denominator.
  • Not simplifying the radical (√20 should be 2√5).
  • Keeping a negative time or length in a real-world answer.