Study guides & original practice
Solving quadratic equations: choose a method, then check
A quadratic equation contains a squared variable as its highest power. Put it in the form ax² + bx + c = 0, with a nonzero, before choosing a solving method.
Worked example: factor a simple quadratic
Solve x² − 7x + 12 = 0. Two numbers that multiply to 12 and add to −7 are −3 and −4. Therefore (x − 3)(x − 4) = 0, giving x = 3 or x = 4.
Check both in the original equation: 9 − 21 + 12 = 0 and 16 − 28 + 12 = 0. Factoring rewrites the expression; setting each factor equal to zero finds the solutions. That step depends on the product being zero.
Move every term before factoring
For x² = 5x + 14, subtract 5x and 14 from both sides. Then x² − 5x − 14 = 0 becomes (x − 7)(x + 2) = 0. The solutions are 7 and −2.
A frequent mistake is treating x² and 5x as like terms. They are different powers and cannot be combined that way. Keep each power in its own term, and carry signs through the rearrangement.
When factoring is awkward
Use the quadratic formula, x = [−b ± √(b² − 4ac)] / (2a). For x² + 2x − 1 = 0, a = 1, b = 2 and c = −1. The result is [−2 ± √8] / 2 = −1 ± √2.
The two approximate values are 0.414 and −2.414. Keep the radical form when an exact answer is requested. The expression under the square root is the discriminant: positive gives two distinct real roots, zero gives one repeated real root, and negative gives no real roots.
For background methods, see OpenStax's quadratic-equations lesson. These worked numbers and practice exercises are original.
Try four algebra questions
1. Solve x² − 9x + 20 = 0.
Reveal the worked answer
(x − 4)(x − 5) = 0, so x = 4 or 5. Their sum is 9 and product is 20; substituting either returns zero.
2. Solve 2x² − 8x = 0.
Reveal the worked answer
Factor 2x(x − 4) = 0. Solutions are 0 and 4. Dividing both sides by x without considering x = 0 would lose a valid solution.
3. Solve x² + 6x + 9 = 0.
Reveal the worked answer
(x + 3)² = 0, so x = −3 is a repeated root. The discriminant is 36 − 36 = 0.
4. How many real solutions does x² + 4 = 0 have?
Reveal the worked answer
None. A real square cannot equal −4. If working over complex numbers, the solutions are 2i and −2i; check which number system the question requests.
What to do after a wrong answer
Expand your factors to check that they reproduce the original expression. If the factors are right but the roots have the wrong signs, solve each linear factor separately. If using the formula, write a, b and c with their signs before substituting and put the entire numerator over 2a.
Check each proposed solution in the original equation, not only your rearranged version. For word problems, also check the meaning: a negative length can be an algebraic root without being a valid length. Make a practice card from the step you missed, then solve a different problem using the same method.
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