How do you know which method to use to solve a quadratic?

Try factoring first, because when it works it is the fastest route and takes about ten seconds. If no pair of whole numbers multiplies to the constant and adds to the middle coefficient, go straight to the formula. Completing the square is the method you choose when the question asks for the vertex, the minimum value, or an exact surd answer rather than just the roots.

What does the discriminant tell you?

The discriminant is b squared minus 4ac, and it is the part under the square root in the quadratic formula. If it is positive there are two distinct real roots, if it is zero the two roots have collapsed into one repeated root, and if it is negative there is no real root because you cannot square-root a negative real number. It answers how many solutions there are without solving anything.

Why does a quadratic have two solutions?

Because squaring destroys sign information. Both 3 and minus 3 square to 9, so undoing a square has to admit both possibilities. Every method carries this: completing the square ends in a plus-or-minus, factoring gives two brackets each of which can be the zero, and geometrically a parabola crosses a horizontal line in two places.

What is the binomial theorem for?

It expands a bracket raised to a power without multiplying it out. To expand (1 + x) to the tenth power by hand you would multiply ten brackets; the theorem gives any single term directly, so you can find the coefficient of x cubed in one line. The coefficients count how many ways you can choose x from each bracket, which is why they are the combination numbers from Pascal's triangle.

Can every equation be solved exactly?

No. Quadratics always have a formula, cubics and quartics have formulas nobody uses, and from the fifth degree upwards no general formula in radicals exists. When an equation has no exact route, you rearrange it into the form x equals g(x) and iterate: feed a guess in, take the output as the next guess, and repeat until the digits stop moving. That is an answer to as many decimal places as you need, not an approximation of a lesser kind.

Quadratics and Polynomials

A quadratic is any equation where the highest power of the unknown is two, and the reason it gets a chapter of its own is that it is the last family of equations that always yields to a formula you can memorise. Everything above it — cubics, quartics, quintics — either factorises by luck or gets solved numerically.

There are three ways to solve one, and they are not interchangeable. Each answers a slightly different question, and knowing which to reach for is most of the skill.

How do you solve a quadratic by factoring?

Get everything on one side, split it into two brackets, and set each bracket to zero in turn. x² + 5x + 6 = 0 factors to (x + 2)(x + 3) = 0, so x = −2 or x = −3.

The search is for two numbers that multiply to the constant and add to the middle coefficient — here 2 and 3, because 2 × 3 = 6 and 2 + 3 = 5. The app's lesson on solving a quadratic by factoring drills exactly that search until it is automatic, since it is the step people slow down on.

Setting each bracket to zero works because of a property of numbers that is so familiar it goes unnoticed: if two things multiply to zero, at least one of them is zero. Nothing else in arithmetic behaves this way — two numbers can multiply to 12 in endless ways, but they can only multiply to 0 by one of them being 0. That is why the whole method insists on getting the equation to equal zero first.

What do you do when it will not factor?

Use the formula: x = (−b ± √(b² − 4ac)) / 2a, which solves every quadratic whether it factors or not. It is worth learning as one object rather than four symbols, and the lesson on the quadratic formula takes it apart term by term so the ± and the 2a stop looking arbitrary.

2x² + 3x − 4 = 0 → x = (−3 ± √(9 + 32)) / 4 = (−3 ± √41) / 4.

Most real quadratics do not factor. Whole-number factorisations are a small and lucky subset, which is why textbook exercises are full of them and physics problems almost never are.

Where does the quadratic formula come from?

It is completing the square, carried out once on the general equation instead of separately on every problem. That is the entire origin story, and deriving the quadratic formula walks the derivation slowly enough that the formula becomes something you could rebuild rather than something you have to trust.

Completing the square rewrites x² + 6x − 1 as (x + 3)² − 10: halve the 6 to get the 3 inside the bracket, then subtract the 9 the bracket brought in with it, along with the original −1. Completing the square is the lesson that makes the halving step feel inevitable rather than magical.

Why halve? Because (x + p)² expands to x² + 2px + p². The middle term always comes out as twice p, so to hit a middle term of 6x you need p = 3. The square you are completing is a literal one — a p-by-p corner tile that finishes off an L-shaped region — and that picture is why the technique is a thousand years older than the notation it is written in.

The rewritten form is worth more than the roots. (x + 3)² − 10 says at a glance that the minimum value is −10, reached at x = −3, because a squared bracket is never negative and is zero exactly once.

How many solutions does a quadratic have, and can you tell before solving?

Yes — compute b² − 4ac, the discriminant, and you know the answer without finishing the calculation. Positive means two distinct roots, zero means one repeated root, negative means no real root at all. The discriminant and the nature of the roots is the lesson for this, and it is the single most useful thing on the page for exam questions phrased as "show that this equation has no real solutions".

The same quantity answers a question that looks unrelated: whether a quadratic expression is ever negative. If a is positive and the discriminant is negative, the parabola sits entirely above the axis, so the expression is always positive for every real x — a fact used constantly in proofs and in checking that a model cannot produce nonsense.

What do the two roots add up to?

They add to −b/a and multiply to c/a, and you can read both off the equation without solving it. The sum and product of the roots shows why: expanding a(x − α)(x − β) and matching it against ax² + bx + c forces those two relationships.

This turns out to be a construction tool. If you know what you want the roots to be, you can build the equation backwards, which is what forming quadratic equations is about — asked for an equation with roots 2 and −5, you write x² − (sum)x + (product) = x² + 3x − 10 without touching a formula.

It scales up too. A cubic has three roots, and they satisfy three matching relationships; roots and coefficients of a cubic extends the pattern, and transforming the roots of a polynomial shows the trick that follows from it: to find the equation whose roots are double the old ones, you never need the old roots at all.

How do you solve a quadratic inequality?

Find the roots first, then decide which side of them you want. For x² − x − 6 < 0, the roots are −2 and 3, and an upward parabola dips below the axis only between its roots, so the answer is −2 < x < 3. Solving a quadratic inequality teaches the sketch-first habit, which is the one that survives contact with a hard question.

The tempting shortcut — divide both sides and treat it like a linear inequality — fails because the sign of the thing you divide by depends on x, and dividing by a negative flips the inequality. The sketch sidesteps the whole problem: the curve is either above the axis or below it, and you are simply reading off which stretch of x does what.

The same caution applies to rational inequalities, where the denominator can change sign; multiplying up without knowing that sign is the classic way to produce a confident wrong answer.

When does a line meet, touch or miss a curve?

Substitute the line into the curve, and read the discriminant of the quadratic that comes out. Two solutions means the line cuts the curve twice, one means it is a tangent, none means it misses entirely — when a line meets, touches or misses a curve is the lesson that connects those three algebraic outcomes to the three pictures.

Getting there needs the substitution step itself, which is the subject of a linear equation paired with a quadratic: rearrange the linear equation for one variable, put it into the quadratic, and solve the single-variable equation that results. The tangency case is the one worth rehearsing, because "the line is a tangent" is a geometric sentence whose algebraic translation is exactly "the discriminant is zero".

Equations that turn into quadratics

A great many equations that look nothing like a quadratic become one after a single rearrangement. Clearing a denominator turns a fractional equation into a quadratic, and squaring both sides does the same for a equation with a square root.

Both moves can invent solutions that were never there. Squaring is the dangerous one: x = −2 is false, but square both sides and x² = 4 is true. The squared equation is a weaker statement than the one you started with, so it admits extra answers. This is why radical equations always end with a check — substituting each candidate back is not diligence, it is part of the method.

What is the binomial theorem?

It is a way to expand a bracket raised to a power without multiplying it out. (1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵, and the coefficients 1, 5, 10, 10, 5, 1 are the row of Pascal's triangle you would expect. The binomial theorem explains where they come from rather than asking you to memorise the triangle.

Multiplying out (1 + x)⁵ means choosing either the 1 or the x from each of five brackets and adding up every possible outcome. A term in x² arises exactly when you picked the x from two brackets and the 1 from the other three — and there are ten ways to choose which two. That is why the coefficients are the combination numbers: they are literally counting choices.

The theorem then extends past whole-number powers, which is where it stops being an algebra convenience and becomes an approximation tool. The binomial series for a negative or fractional index turns things like √(1 + x) and 1/(1 + x) into infinite series that are accurate for small x, and expanding (a + bx) to a rational power handles the case where the bracket does not conveniently start with a 1 — you factor the a out first, and the condition for validity comes along with it.

How do you factorise a cubic?

Find one root by trial, then divide it out. If substituting x = 2 into a polynomial gives zero, then (x − 2) is a factor of it, and dividing reduces a cubic to a quadratic you already know how to finish. The factor and remainder theorem is the statement that makes this legitimate, and it also tells you what the leftover means when the substitution does not give zero: the value you get is exactly the remainder.

Which roots to try is not a guessing game — if the polynomial has whole-number coefficients, any whole-number root must divide the constant term, so a cubic ending in −6 has only ±1, ±2, ±3, ±6 worth testing.

What if there is no exact answer?

Rearrange the equation into the shape x = g(x), start from a rough guess, and feed each output back in as the next input. The values usually settle, and the digits that stop changing are correct digits. Solving equations by iteration covers the method and, importantly, when it fails — some rearrangements of the same equation converge quickly, others fly off to infinity, and which is which is not obvious in advance.

This matters more than it first appears. Beyond the fourth degree there is no general formula in radicals, so iteration is not a fallback for the lazy: for most polynomial equations that arise in practice, it is the method.

The mistakes worth naming

Solving before setting it to zero. (x + 2)(x + 3) = 2 tells you nothing about the brackets. The zero-product property needs the zero.
Losing the negative root. √9 is 3, but "x² = 9" has two answers. The square root symbol and the act of undoing a square are different things.
Forgetting to subtract what completing the square added. (x + 3)² is not x² + 6x; it carries a spare 9.
Trusting a squared equation. Every radical equation ends with substituting the answers back in.

Learn this properly in the app

Math Challenge teaches quadratics as an illustrated ladder rather than a formula sheet — each lesson linked above is a short screen with a picture, a worked example and try-it problems, and they sit inside a catalog of 800+ lessons running from counting to calculus. The natural next step from here is graphs of equations and coordinate geometry, which is the same parabola seen from the other side, and functions and rational functions, where completing the square reappears as a graph transformation.

Your turn

Three to try — tap what you get.

Factor x² − 9

The roots of (x − 2)(x + 5) = 0

x² + 6x + 9 =

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