The Quadratic Formula & Nature of the Roots

Use quadratic formula and analyze discriminant. Grade 11 Math 20-1 curriculum - exam essential.

Lesson 1.9 of Quadratic Functions & Equations in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• The Quadratic Formula. Derive the quadratic formula from completing the square, then apply it to four worked examples including equations with rational, irrational, and non-real solutions.
• The Discriminant — Predicting the Roots. Use the discriminant to determine the number and type of roots of any quadratic equation without solving it, and connect the discriminant to the number of x-intercepts on the parabola.
• Working Backwards — From Roots to Discriminant. Given the roots of a quadratic equation, determine the sign of the discriminant, and find unknown parameters that produce a specific type of root.
• Choosing the Best Method. Compare all four solving methods — factoring, common factor, square root principle, and the quadratic formula — and use a decision flowchart to pick the most efficient approach.

Inside the lesson — a free preview

Where Does the Formula Come From?

The quadratic formula is what you get when you complete the square on the general equation ax^2 + bx + c = 0. Instead of redoing all that algebra every time, the formula does it for you.

The Quadratic Formula. For any quadratic equation in the form ax^2 + bx + c = 0, the solutions are:

x = (-b ± √(b^2 - 4ac))/(2a)

where a ≠ 0.

When to Use the Quadratic Formula. Use the quadratic formula whenever factoring is too hard or fails. The formula always works — for any quadratic equation, regardless of whether the roots are rational, irrational, or non-real.

Worked example: Worked Examples — Applying the Formula

Example 1a — Solve 3x^2 + 11x = -10. Step 1 — Set equal to zero.

3x^2 + 11x + 10 = 0

Step 2 — Identify a, b, c.

a = 3, b = 11, c = 10

Step 3 — Substitute into the formula.

x = (-11 ± √(11^2 - 4(3)(10)))/(2(3))

Step 4 — Simplify under the radical.

11^2 = 121, 4(3)(10) = 120, 121 - 120 = 1

x = (-11 ± √(1))/(6) = (-11 ± 1)/(6)

Step 5 — Split into two solutions.

x = (-11 + 1)/(6) = (-10)/(6) = -(5)/(3) and x = (-11 - 1)/(6) = (-12)/(6) = -2

Answer: x = -(5)/(3) and x = -2

Example 1b — Solve x^2 - 4x + 1 = 0. Step 1 — Already equal to zero.

Step 2 — Identify a, b, c.

a = 1, b = -4, c = 1

Step 3 — Substitute into the formula.

x = (-(-4) ± √((-4)^2 - 4(1)(1)))/(2(1)) = (4 ± √(16 - 4))/(2) = (4 ± √(12))/(2)

Step 4 — Simplify the radical.

√(12) = √(4 · 3) = 2√(3), so x = (4 ± 2√(3))/(2)

Step 5 — Reduce by dividing every term by 2.

Answer: x = 2 + √(3) and x = 2 - √(3)

Common Mistake — Negative b Values. When b is negative, -b becomes positive. Use brackets carefully:

b = -4 → -b = -(-4) = +4

Forgetting the double negative is one of the most common errors with the quadratic formula.

Example 1c — Solve (2d + 3)(d - 2) = (d + 9)(d - 3). Step 1 — Expand both sides.

LHS: (2d + 3)(d - 2) = 2d^2 - 4d + 3d - 6 = 2d^2 - d - 6

RHS: (d + 9)(d - 3) = d^2 - 3d + 9d - 27 = d^2 + 6d - 27

Step 2 — Move everything to one side.

2d^2 - d - 6 - d^2 - 6d + 27 = 0

d^2 - 7d + 21 = 0

Step 3 — Identify a, b, c.

a = 1, b = -7, c = 21

Step 4 — Substitute into the formula.

d = (7 ± √(49 - 84))/(2) = (7 ± √(-35))/(2)

Step 5 — Interpret.

The number under the square root is negative. You cannot take the square root of a negative number in the real numbers.

Answer: No real solutions.

Example 1d — Solve 7x^2 + 4x + 2 = 0. Step 1 — Already equal to zero.

Step 2 — Identify a, b, c.

a = 7, b = 4, c = 2

Step 3 — Substitute and simplify.

x = (-4 ± √(16 - 4(7)(2)))/(14) = (-4 ± √(16 - 56))/(14) = (-4 ± √(-40))/(14)

The value under the square root is negative.

Answer: No real solutions.

Notice: When the value under the square root (called the discriminant) is negative, there are no real solutions. We will explore this idea in the next tab.

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