Short answer
How to solve quadratic equations
Try factorising first, because it is fastest when it works. If the quadratic does not factorise neatly, use the formula x = (−b ± √(b² − 4ac)) / 2a. Complete the square when the question asks for it, when you need the vertex, or when you are deriving something.
Check the discriminant first — b² − 4ac tells you how many real roots exist before you start.
Get it into standard form first
Every method assumes ax² + bx + c = 0, with everything on one side and zero on the other. Rearranging first prevents most sign errors.
Identify a, b and c explicitly, including their signs, and write them down. For 3x² − 5x = 2, that is 3x² − 5x − 2 = 0, so a = 3, b = −5, c = −2.
The negative sign belongs to the coefficient. Students who write b = 5 for the equation above will get the wrong answer from the formula and will not be able to see why.
The discriminant tells you what to expect
b² − 4ac, computed before you solve anything, tells you how many real solutions there are.
Positive: two distinct real roots. Zero: one repeated root, and the parabola touches the x-axis. Negative: no real roots, and the solutions are complex.
A perfect square discriminant means the quadratic factorises with rational numbers, which tells you factorising will work. That single check saves the time otherwise spent hunting for factors that do not exist.
Questions frequently ask about the discriminant directly — 'find the values of k for which the equation has equal roots' means set b² − 4ac = 0 and solve for k.
Method one: factorising
Fastest when it works. For x² + bx + c with a = 1, find two numbers that multiply to c and add to b.
For x² + 7x + 12: two numbers multiplying to 12 and adding to 7 are 3 and 4, so it factorises to (x + 3)(x + 4) = 0, giving x = −3 and x = −4.
When a is not 1, multiply a by c and find two numbers that multiply to ac and add to b, then split the middle term and factorise in pairs. For 3x² − 5x − 2, ac = −6, and the numbers are −6 and 1: 3x² − 6x + x − 2 = 3x(x − 2) + 1(x − 2) = (3x + 1)(x − 2).
The sign of the constant tells you what to look for. Positive c means both factors have the same sign; negative c means they differ.
If you cannot find the factors within about thirty seconds, use the formula. Hunting is where time is lost in an exam.
Method two: the quadratic formula
x = (−b ± √(b² − 4ac)) / 2a. It always works, and it is the fallback when factorising does not.
Substitute carefully and use brackets, particularly when b is negative. For 3x² − 5x − 2 = 0: x = (5 ± √(25 + 24)) / 6 = (5 ± 7) / 6, giving x = 2 and x = −1/3.
The whole of −b is over 2a, not just the root. Writing the fraction bar across the entire numerator prevents the commonest formula error.
Leave the answer in surd form unless the question asks for decimals, and if it asks for decimals, give the number of places specified.
Method three: completing the square
Rewrite ax² + bx + c in the form a(x + p)² + q. Halve the coefficient of x, square it, add and subtract it.
For x² + 6x + 5: half of 6 is 3, and 3² is 9, so x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. Setting that to zero gives (x + 3)² = 4, so x + 3 = ±2 and x = −1 or −5.
Where a is not 1, factor it out of the first two terms first: 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2((x + 2)² − 4) + 3 = 2(x + 2)² − 5.
This is the method to use when the question asks for the vertex, the minimum or maximum value, or the line of symmetry, because the completed form gives all three directly: the vertex of a(x + p)² + q is at (−p, q).
It is also how the quadratic formula is derived, which is why questions ask for it even when factorising would be quicker.
Which method when
Read the question first. If it names a method, use it — marks are for the method as well as the answer.
| Situation | Method |
|---|---|
| Discriminant is a perfect square | Factorise |
| Question says 'solve' | Factorise if quick, formula otherwise |
| Question asks for the vertex, minimum or maximum | Complete the square |
| Question asks for exact solutions in surd form | Formula or completing the square |
| Question asks for the number of roots only | Discriminant, no need to solve |
| Question specifies a method | That method, even if another is faster |
| Coefficients are awkward or non-integer | Formula |
Checking your answer
Substitute both roots back into the original equation. It takes twenty seconds and it catches sign errors, which are the most common mistake in all three methods.
There is a faster check for factorised answers: the two roots should sum to −b/a and multiply to c/a. For 3x² − 5x − 2, the roots 2 and −1/3 sum to 5/3 which is −b/a, and multiply to −2/3 which is c/a.
If the discriminant was negative and you have produced two real roots, you have made an arithmetic error somewhere.
Sketching the parabola
Questions frequently ask for a sketch after solving, and four pieces of information are enough: the direction, the roots, the y-intercept and the vertex.
The parabola opens upwards if a is positive and downwards if a is negative. The roots are where it crosses the x-axis. The y-intercept is c. The vertex comes from the completed square, or from x = −b/2a.
Label all of them on the sketch. An unlabelled curve of the right shape earns very little; a roughly drawn curve with the roots, intercept and vertex marked earns most of the marks.
Where quadratics appear
Beyond the algebra questions: projectile motion in mechanics, where the height is a quadratic in time and the roots are when the object is at ground level; optimisation problems, where the vertex is the maximum or minimum; and any problem producing an area or a product with a constraint.
In those contexts, check whether both roots make physical sense. A negative time or a negative length is mathematically valid and physically not, and the question usually expects you to discard it and say why.
Simultaneous equations with one linear and one quadratic reduce to a single quadratic by substitution, and the discriminant then tells you whether the line meets the curve twice, once as a tangent, or not at all.
Quadratics that are not obviously quadratic
Some equations become quadratic after a substitution, and recognising them is what the question is testing.
Quartics in disguise: x⁴ − 5x² + 4 = 0 becomes u² − 5u + 4 = 0 with u = x². Solve for u, then remember to solve back for x — each positive value of u gives two values of x, and a negative u gives none.
Exponentials: 2^(2x) − 5·2^x + 4 = 0 becomes u² − 5u + 4 = 0 with u = 2^x. Since 2^x is always positive, discard any non-positive root of u before solving back.
Equations with surds often become quadratic after squaring, and squaring can introduce solutions that do not satisfy the original. Always substitute the answers back and discard the ones that fail.
Trigonometric equations: 2sin²x + 3sin x − 2 = 0 is quadratic in sin x. Solve for sin x, then find all the angles in the given range, which is where marks are usually lost rather than in the quadratic.
The errors that lose marks
Six, and the first two account for most of them.
- Sign errors in b, particularly when substituting a negative b into the formula.
- Putting only −b over 2a rather than the whole numerator.
- Forgetting the ± and giving only one root.
- Not rearranging to standard form before identifying a, b and c.
- Rounding early and losing accuracy in the final answer.
- Giving a decimal when the question asked for an exact answer in surd form.
Frequently asked questions
Which method should I use?
Factorise if the discriminant is a perfect square, because it is fastest. Use the formula otherwise. Complete the square when the question asks for it or when you need the vertex, minimum or maximum.
What does the discriminant tell me?
b² − 4ac gives the number of real roots: positive means two, zero means one repeated root, negative means none. A perfect square also tells you the quadratic will factorise with rational numbers.
What if the quadratic will not factorise?
Use the formula. If you cannot find factors within about thirty seconds, stop hunting — that is where exam time is lost.
How do I complete the square when a is not 1?
Factor a out of the first two terms only, complete the square inside the bracket, then expand the a back through and simplify the constant.
Should I leave answers as surds or decimals?
As surds unless the question asks for decimals, and if it does, to exactly the accuracy specified. Rounding early loses marks even when the method is right.
How do I find the vertex?
From the completed square form a(x + p)² + q, the vertex is at (−p, q). Alternatively the x-coordinate is −b/2a, and substituting gives the y-coordinate.