Solving Quadratic Equations: Simple Cases, Factoring and Zero-Product Property
A quadratic equation has the form ax2 + bx + c = 0 with a ≠ 0, and it can often be solved by taking square roots, by factoring with the zero-product property, or by other methods. In brief: x2 = 49 gives x = ±7, x2 − 5x = 0 factors as x(x − 5) = 0 so x = 0 or 5, and x2 − 5x + 6 = 0 factors as (x − 2)(x − 3) = 0 so x = 2 or 3.
A quadratic equation has up to two real solutions. Which method is fastest depends on the shape of the equation, and recognising that shape is the main skill. This guide explains three simple cases, the zero-product property, factoring in the general case and a quick check by substitution, with a table that tells you which approach to choose.
How do you solve the simple cases?
Case 1: ax2 = c (no linear term)
Isolate x2 and take square roots. For 3x2 = 75: x2 = 25, so x = 5 or x = −5. If the right side is negative, such as x2 = −4, there is no real solution, because squares are not negative. If x2 = 0, the only solution is x = 0.
Case 2: ax2 + bx = 0 (no constant term)
Factor out x: x(ax + b) = 0. The solutions are x = 0 and x = −b/a. For 4x2 − 12x = 0: 4x(x − 3) = 0, so x = 0 or 3.
Case 3: (x − p)2 = q
Take square roots: x − p = ±√q, so x = p ± √q. For (x − 3)2 = 10: x = 3 ± √10. For (2x + 1)2 = 9: 2x + 1 = ±3, so x = 1 or x = −2.
What is the zero-product property?
If a product of two numbers is zero, at least one of them is zero: A · B = 0 implies A = 0 or B = 0. This is true only for zero. For a product equal to 6, you cannot conclude that one factor is 6. The property is the reason why we move everything to one side and make the other side zero before factoring.
How do you solve by factoring step by step?
- Bring the equation to the standard form ax2 + bx + c = 0.
- Factor the left side.
- Set each factor equal to zero.
- Solve the linear equations.
- Check by substitution.
Example 1: x2 − 5x + 6 = 0
Two numbers with product 6 and sum −5: −2 and −3. So (x − 2)(x − 3) = 0 and x = 2 or x = 3. Check: 4 − 10 + 6 = 0 ✓ and 9 − 15 + 6 = 0 ✓.
Example 2: x2 + 2x = 15
First move: x2 + 2x − 15 = 0. Numbers with product −15 and sum 2: 5 and −3. So (x + 5)(x − 3) = 0 and x = −5 or 3. Do not write (x)(x + 2) = 15 and conclude x = 15.
Example 3: 6x2 + x − 2 = 0
a · c = −12, and the numbers 4 and −3 give sum 1. Split: 6x2 + 4x − 3x − 2 = 2x(3x + 2) − (3x + 2) = (3x + 2)(2x − 1). So x = −2/3 or x = 1/2.
Example 4: double root
x2 − 6x + 9 = 0 gives (x − 3)2 = 0, so x = 3 is a double root. The parabola touches the x-axis once.
Example 5: a difference of squares
4x2 − 25 = 0 gives (2x − 5)(2x + 5) = 0, so x = ±5/2.
Which method fits which equation?
| Shape of the equation | Method | Example and solutions |
|---|---|---|
| ax2 = c | Square root | x2 = 16: x = ±4 |
| ax2 + bx = 0 | Factor out x | x2 − 7x = 0: x = 0, 7 |
| (x − p)2 = q | Square root | (x − 1)2 = 9: x = 4, −2 |
| Integer factors easy to see | Factoring | x2 + x − 12 = 0: x = 3, −4 |
| Difficult coefficients | Quadratic formula | x2 − 3x − 5 = 0: x = (3 ± √29)/2 |
What if the equation does not factor?
Not every quadratic equation factors with integers. In that case use completing the square or the quadratic formula. A quick test: if the discriminant b2 − 4ac is a perfect square, the equation factors over the integers or rational numbers. For x2 − 3x − 5 the discriminant is 29, which is not a perfect square, so factoring fails.
Where do these methods come from?
Babylonian clay tablets from around 2000 BCE solve problems that correspond to quadratic equations. Al-Khwarizmi’s ninth-century book on algebra gave systematic rules for all types, using geometric arguments. The zero-product property was used implicitly for centuries, and the idea of factoring as the main solution method became standard with the use of symbolic algebra from the sixteenth century.
Where are quadratic equations used today?
Questions such as “when does a ball hit the ground” lead to quadratic equations. A rectangle whose length is 3 m more than its width and whose area is 40 m2 satisfies x(x + 3) = 40, so x2 + 3x − 40 = 0, which factors as (x + 8)(x − 5) = 0. The width is x = 5 m, and the negative solution is rejected because lengths are positive. Engineering, finance and physics give many similar equations.
Which mistakes are common?
Forgetting the negative root
x2 = 49 has two solutions, 7 and −7.
Dividing both sides by x
From x2 = 5x, dividing by x loses x = 0. Factor instead.
Not setting the equation to zero
The zero-product property works only when the product equals 0.
Not rejecting solutions that do not fit the context
A negative length or time must be rejected in word problems.