Quadratic Equations • Topic 5 of 5

Quadratic Formula

The quadratic formula solves any quadratic ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a). Substitute a, b and c with their signs, compute the discriminant b² − 4ac, then evaluate both roots. The discriminant also tells you how many real solutions there are: positive gives two, zero gives one (a repeated root), and negative gives none (no real solutions). Use the formula when factoring is hard or the roots are not integers. On the Digital SAT the built-in calculator handles the arithmetic, but knowing the formula and discriminant is essential for setup.

✅ Solved examples

1. Solve x² + 5x + 6 = 0 with the formula.
a=1, b=5, c=6; discriminant = 25 − 24 = 1; x = (−5 ± 1)/2 = −2 or −3.
2. How many real solutions does x² + 2x + 5 = 0 have?
Discriminant = 4 − 20 = −16 < 0, so no real solutions.
3. Solve x² − 4x + 4 = 0.
Discriminant = 16 − 16 = 0; one root x = 4/2 = 2.
4. Find the discriminant of 2x² + 3x − 2 = 0.
b² − 4ac = 9 − 4(2)(−2) = 9 + 16 = 25.

✏️ Practice — try these, take hints as needed

1. How many real solutions does x² − 6x + 9 = 0 have?
Compute b² − 4ac.
36 − 36 = 0.
D = 0 means…
One (repeated) real solution.
2. How many real solutions does x² + x + 1 = 0 have?
Discriminant = 1 − 4.
−3 < 0.
—
None.
3. Find the discriminant of x² − 5x + 6 = 0.
b² − 4ac.
25 − 24.
—
1.
4. Solve x² + 6x + 5 = 0 with the formula.
D = 36 − 20 = 16, √16 = 4.
x = (−6 ± 4)/2.
—
x = −1 or x = −5.
5. How many real solutions does x² − 2x − 3 = 0 have?
D = 4 + 12 = 16.
Positive discriminant.
—
Two real solutions.

📝 Topic test — 8 questions

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