6.3 The Quadratic Formula, the Discriminant & Quadratic Modeling

6.3 The Quadratic Formula, the Discriminant & Quadratic Modeling

The quadratic formula is the universal method: it works whether or not the quadratic factors, whether or not the coefficients are integers, and whether or not real solutions exist. Because test 025 provides no reference sheet, you must have it memorized.

For ax² + bx + c = 0 with a ≠ 0, x = ( −b ± √(b² − 4ac) ) / (2a)

The formula is what completing the square produces when carried out on the general form, which is why the two methods always agree.

Substitution mechanics

Write the equation in standard form first, then list a, b, and c with their signs before substituting. Most errors are bookkeeping, not algebra.

Solve 2x² − 7x − 15 = 0 a = 2, b = −7, c = −15 Discriminant: b² − 4ac = (−7)² − 4(2)(−15) = 49 + 120 = 169 x = (7 ± √169)/(2 · 2) = (7 ± 13)/4 x = 20/4 = 5 or x = −6/4 = −3/2

Three sign checkpoints in that computation:

1. −b = −(−7) = +7. A negative b becomes positive in the numerator.
2. b² = (−7)² = 49, always positive regardless of b's sign.
3. −4ac = −4(2)(−15) = +120. Two negatives in the product make the term positive, which is why the discriminant grew rather than shrank.

Because the discriminant was the perfect square 169, this equation also factors — as (x − 5)(2x + 3) = 0. A perfect-square discriminant always signals that factoring was available.

The discriminant

The expression under the radical, D = b² − 4ac, answers "how many real solutions?" without finishing the problem.

+---------------------------------------------------------------------------+
|  D > 0   ->  TWO distinct real solutions   parabola crosses x-axis twice  |
|              (D a perfect square -> the solutions are rational)           |
|  D = 0   ->  ONE repeated real solution    vertex sits ON the x-axis      |
|  D < 0   ->  NO real solutions             parabola never touches x-axis  |
+---------------------------------------------------------------------------+

For x² − 6x + 9 = 0: D = 36 − 36 = 0, so there is one repeated root. Indeed (x − 3)² = 0 gives x = 3, and the vertex (3, 0) sits on the axis. For 2x² + 3x + 5 = 0: D = 9 − 40 = −31 < 0, so there are no real solutions and the parabola lies entirely above the x-axis.

The graphical reading is what makes the discriminant a Competency 2 skill rather than trivia: it tells you whether the modeled quantity ever reaches zero.

Modeling and root selection

A ball is thrown so that its height in feet is h(t) = −16t² + 48t + 6. When does it hit the ground?

Ground level means h = 0, so −16t² + 48t + 6 = 0. Multiply by −1 for convenience: 16t² − 48t − 6 = 0, and divide by 2: 8t² − 24t − 3 = 0.

• a = 8, b = −24, c = −3
• D = 576 − 4(8)(−3) = 576 + 96 = 672
• √672 ≈ 25.92
• t = (24 ± 25.92)/16 → t ≈ 49.92/16 ≈ 3.12 s or t ≈ −1.92/16 ≈ −0.12 s

Reject the negative root. Time before the throw has no meaning, so the ball lands at about 3.12 seconds. Items always include the negative root as an option, and sometimes ask explicitly why it is rejected — the answer is contextual, not algebraic. The root is a perfectly valid solution to the equation; it is the model that excludes it.

The same reasoning governs geometry models:

A rectangular garden is 4 feet longer than it is wide and has an area of 96 square feet. Find its dimensions. w(w + 4) = 96 → w² + 4w − 96 = 0 → (w + 12)(w − 8) = 0 → w = −12 or w = 8. Width cannot be negative, so w = 8 ft and length = 12 ft. Check: 8 · 12 = 96 ✓

Reaching a target value rather than zero

Many items ask when a quantity equals some nonzero value. Set the function equal to that value and re-standardize.

When is the ball at height 30 feet? −16t² + 48t + 6 = 30 → −16t² + 48t − 24 = 0 → divide by −8: 2t² − 6t + 3 = 0. D = 36 − 24 = 12, so t = (6 ± √12)/4 = (6 ± 3.46)/4, giving t ≈ 0.63 s and t ≈ 2.37 s.

Both roots are meaningful here — the ball passes 30 feet on the way up and again on the way down. Recognizing when to keep both roots and when to discard one is the interpretive skill being assessed. A useful check: the two times should be symmetric about the vertex time t = −48/(2 · −16) = 1.5, and indeed (0.63 + 2.37)/2 = 1.5. ✓

Method selection under time pressure

Given about three minutes per item, choose deliberately:

• If the constant term factors obviously with a = 1, factor — it is fastest.
• If the equation has the form (something)² = number, use the square root method.
• If you need the vertex as well as the roots, complete the square.
• If nothing is clean, go straight to the formula. Computing the discriminant first is worthwhile: if D < 0 you are done immediately, and if D is a perfect square you can switch back to factoring.