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Understanding Quadratic Equations for NECO

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a is not zero. NECO, WAEC, and JAMB all test this topic heavily, and almost every question comes down to one of three solving methods. This guide explains each method in plain language, works through a full example for each, and covers the mistakes that most often cost marks.

1. Factorization

Factorization works when the equation splits neatly into two brackets. The method: find two numbers that multiply together to give a × c, and add together to give b. Once found, use them to split the middle term (bx) into two parts, then factor by grouping.

Question: Solve x² + 2x - 15 = 0
  1. Identify a = 1, b = 2, c = -15.
  2. Look for two numbers that multiply to -15 and add to 2: that's 5 and -3.
  3. Rewrite as (x + 5)(x - 3) = 0.
  4. Set each bracket to zero: x + 5 = 0 or x - 3 = 0.
  5. Solve each: x = -5 or x = 3.
Answer: x = -5 or x = 3

2. Completing the square

This method is used when factorization isn't clean, when there's no obvious pair of numbers that multiply and add correctly. Instead, you rewrite one side of the equation as a perfect square.

Question: Solve x² + 6x + 4 = 0 by completing the square.
  1. Move the constant to the other side: x² + 6x = -4.
  2. Take half of the coefficient of x (which is 6), giving 3, then square it: 3² = 9.
  3. Add 9 to both sides: x² + 6x + 9 = -4 + 9, which simplifies to x² + 6x + 9 = 5.
  4. The left side is now a perfect square: (x + 3)² = 5.
  5. Take the square root of both sides: x + 3 = ±√5.
  6. Solve for x: x = -3 ± √5.
Answer: x = -3 + √5 or x = -3 - √5 (leave in surd form unless the question asks for a decimal)

3. The quadratic formula

The quadratic formula works for every quadratic equation, no matter how messy the numbers are: x = (-b ± √(b² - 4ac)) / 2a. If factorization looks difficult and completing the square feels slow, go straight to this formula.

Question: Solve 2x² + 5x - 3 = 0 using the quadratic formula.
  1. Identify a = 2, b = 5, c = -3.
  2. Calculate the part under the square root first: b² - 4ac = 5² - 4(2)(-3) = 25 + 24 = 49.
  3. Take the square root: √49 = 7.
  4. Substitute into the formula: x = (-5 ± 7) / 4.
  5. Solve both cases: x = (-5 + 7)/4 = 2/4 = 0.5, or x = (-5 - 7)/4 = -12/4 = -3.
Answer: x = 0.5 or x = -3

How to choose the right method quickly

Under exam time pressure, don't spend too long deciding. A practical approach: glance at the equation for 5 to 10 seconds to see if factor pairs are obvious. If they are, factorize, it's the fastest method. If nothing obvious appears, or the numbers under the square root (b² - 4ac) don't look like a perfect square, go straight to the quadratic formula rather than wasting time forcing a factorization that isn't there.

Common mistakes to avoid

The most frequent error is a sign mistake when substituting into the quadratic formula, especially when b or c is already negative. Write out b² - 4ac as its own separate step before touching the rest of the formula, this isolates the part most likely to go wrong. A second common mistake is stopping after finding one root and forgetting the ± gives two possible answers, both of which are usually needed.

Frequently asked questions

What if b² - 4ac comes out negative?
That means the equation has no real solutions, only complex ones. At WAEC and NECO level, this usually signals a substitution error, so double-check your values of a, b, and c first.

Do I always need to simplify surd answers?
Simplify where possible (for example, √8 becomes 2√2), but leave the answer in surd form unless the question specifically asks for a decimal approximation.

Can every quadratic equation be factorized?
No. Many quadratic equations only have solutions in surd or decimal form, which is exactly why completing the square and the quadratic formula exist as backup methods.

Exam tip: If you spend more than 30 seconds searching for factor pairs without success, switch to the quadratic formula immediately. It always works, and it's faster than continuing to guess at factor pairs under time pressure.
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