4.2 Quadratic equations (factoring, quadratic formula)

Quadratic Equations

A quadratic equation has the standard form ax^2 + bx + c = 0, where a ≠ 0. Because the highest power is 2, a quadratic can have two, one, or zero real solutions. Those solutions — also called roots or zeros — are the x-values that make the equation true. The PERT tests three tools: solving by factoring, the quadratic formula, and the discriminant. Always rewrite the equation so that one side equals zero before you begin.

Solving by factoring and the zero-product property

The zero-product property states that if A · B = 0, then A = 0 or B = 0. So once a quadratic is factored, set each factor equal to zero and solve the two simple equations.

Example 1. Solve x^2 + 2x - 15 = 0. Factor the trinomial: two numbers that multiply to -15 and add to 2 are 5 and -3, giving (x + 5)(x - 3) = 0. Apply the zero-product property:

• x + 5 = 0 → x = -5
• x - 3 = 0 → x = 3

The solutions are x = -5 and x = 3. Verify the first: (-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0. Correct.

Example 2. Solve x^2 = 6x. Do not divide both sides by x — that would lose a solution. Instead move everything to one side: x^2 - 6x = 0, factor out the GCF x(x - 6) = 0, so x = 0 or x = 6.

The quadratic formula

When a quadratic does not factor with whole numbers, use the quadratic formula, which solves any equation written in standard form:

x = (−b ± √(b² − 4ac)) / (2a)

Identify a, b, and c carefully, keeping their signs, and then substitute.

Example 3. Solve x^2 - 5x + 6 = 0. Here a = 1, b = -5, c = 6.

x = (−(−5) ± √((−5)² − 4·1·6)) / (2·1) x = (5 ± √(25 − 24)) / 2 x = (5 ± √1) / 2 = (5 ± 1) / 2

So x = 6/2 = 3 or x = 4/2 = 2. This one also factors as (x - 2)(x - 3), which confirms the answer.

Example 4. Solve 2x^2 + 3x - 2 = 0, where a = 2, b = 3, c = -2.

x = (−3 ± √(3² − 4·2·(−2))) / (2·2) x = (−3 ± √(9 + 16)) / 4 x = (−3 ± √25) / 4 = (−3 ± 5) / 4

So x = 2/4 = 1/2 or x = -8/4 = -2.

The discriminant

The expression under the radical, b² − 4ac, is the discriminant. Its sign tells you how many real solutions exist before you finish the arithmetic.

Example 5. How many real solutions does x^2 + 4x + 5 = 0 have? Compute b² − 4ac = 4² − 4·1·5 = 16 − 20 = −4. Because the discriminant is negative, there are no real solutions. Compare with x^2 - 6x + 9 = 0, where b² − 4ac = 36 − 36 = 0, giving exactly one repeated solution, x = 3.

An applied quadratic

Word problems often lead to a quadratic. Set up the equation, move everything to one side, then solve.

Example 6. A rectangular garden is 3 feet longer than it is wide, and its area is 40 square feet. Find the width. Let the width be w; then the length is w + 3. Area gives w(w + 3) = 40, so w^2 + 3w - 40 = 0. Factor: two numbers that multiply to -40 and add to 3 are 8 and -5, giving (w + 8)(w - 5) = 0. Then w = -8 or w = 5. A width cannot be negative, so w = 5 feet and the length is 8 feet. Check: 5 · 8 = 40. Correct.

Quick strategy

• Rewrite in standard form ax² + bx + c = 0 first.
• Try factoring; it is fastest when it works cleanly.
• Use the quadratic formula whenever factoring stalls.
• Discard any answer that makes no physical sense in an applied problem.

Mastering these three tools — factoring, the formula, and the discriminant — covers every quadratic item you are likely to meet on the PERT Mathematics subtest.