➕ Algebra 1 · Quadratics
Solving Quadratics by Factoring: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
If (x − 3)(x + 8) = 0, then x = 3 or x = −8.
✓ True. A product is 0 only if a factor is 0. x − 3 = 0 gives x = 3, and x + 8 = 0 gives x = −8.
From (x − 5)(x − 2) = 10 you can conclude that x − 5 = 10 or x − 2 = 10.
✗ False. The product is 10, not 0, so neither factor has to equal 10. Moving the 10 gives x2 − 7x = 0, so x(x − 7) = 0 and x = 0 or x = 7. Check x = 0: (−5)(−2) = 10.
A quadratic equation has at most two real solutions.
✓ True. The solutions are the x-intercepts of a parabola, and a parabola meets the x-axis at most twice. There can be two solutions, one (a double root) or none.
The equation x2 = 15x has only one solution, x = 15.
✗ False. Dividing by x loses a solution. x2 − 15x = 0 gives x(x − 15) = 0, so x = 0 or x = 15. Both work: 0 = 0 and 225 = 225.
The equation (x + 10)2 = 0 has exactly one solution, x = −10.
✓ True. Both factors are x + 10, so setting them to 0 gives the same value. It is a double root, counted as one solution.
If x2 − 6x + 4 cannot be factored with integers, then x2 − 6x + 4 = 0 has no solution.
✗ False. Not factoring is not the same as no solution. At x = 0 the value is 4 and at x = 1 it is 1 − 6 + 4 = −1, so the graph crosses the x-axis between 0 and 1. The solutions exist but are not integers.
The equation 6x(x − 2) = 0 has two solutions, x = 0 and x = 2.
✓ True. The factor 6 is never 0. The factor x gives x = 0 and the factor x − 2 gives x = 2.
If the graph of y = x2 − 5x − 14 crosses the x-axis at (7, 0) and (−2, 0), then the solutions of x2 − 5x − 14 = 0 are x = 7 and x = −2.
✓ True. The x-intercepts are the x-values where y = 0, which are exactly the solutions of the equation. Check: (x − 7)(x + 2) = x2 − 5x − 14.
A negative solution of an area equation can be used as a width, because it makes the equation true.
✗ False. It solves the algebra, but a width cannot be negative. Always check each solution against the situation and reject any that cannot happen.
The solutions of 4x2 − 25 = 0 are x = 5 and x = −5.
✗ False. 4x2 − 25 = (2x − 5)(2x + 5), so x = 5/2 or x = −5/2. Check x = 5: 4(25) − 25 = 75, not 0.
To solve x2 + 12 = 7x by factoring, first rewrite it as x2 − 7x + 12 = 0. The solutions are x = 3 and x = 4.
✓ True. Subtract 7x from both sides, then factor: (x − 3)(x − 4) = 0. Check in the original: 9 + 12 = 21 = 7(3) and 16 + 12 = 28 = 7(4).
If ab = 12, then a = 12 or b = 12.
✗ False. Only a product of 0 works this way. For example, 3 × 4 = 12, and neither number is 12.
If A × B = 0 and A ≠ 0, then B = 0.
✓ True. Divide both sides by A, which is allowed because A is not zero: B = 0 ÷ A = 0. This is why the zero-product property is true.