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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