Quadratic Equations

Here is a problem. Find a number x such that

Stop and try it. With a bit of guessing, you will quickly land on x = 2 — and indeed, 4 + 12 = 16. The answer is 2.

But here is something strange. Try x = -8. You get 64 + (-48) = 16. That also works. So the equation has two answers, not one.

That is curious. A simple equation like 2x = 6 has exactly one answer. But x^2 + 6x = 16 has two. Why?

The answer is that the squared term — the x^2 — changes the rules. An equation with a squared unknown can have up to two solutions, where a linear equation has exactly one. Such an equation is called a quadratic equation — quad from the Latin quadratus, meaning "square," because the unknown is squared. They are the next step after linear equations in school, and in many ways they are the simplest equations that are genuinely interesting.

A linear equation you can solve in your head. A quadratic equation has exactly two solutions, and the path from the equation to the answer goes through one of the most beautiful pieces of geometry in elementary algebra.

What a quadratic equation looks like

A quadratic equation is any equation you can rearrange into the form

where a, b, and c are numbers, and a \neq 0 (otherwise it isn't quadratic at all — just linear). The Babylonian problem x^2 + 6x = 16 rearranges into x^2 + 6x - 16 = 0, so a = 1, b = 6, c = -16.

The thing on the left, ax^2 + bx + c, is a polynomial of degree two. If you graph y = ax^2 + bx + c, you get a curve called a parabola — the same U-shape you have probably seen for the trajectory of a thrown ball. Solving the equation ax^2 + bx + c = 0 is the same as asking where the parabola crosses the horizontal axis. Every quadratic equation is a parabola asking the same question: where am I zero?

There are exactly three things that can happen.

The parabola can cut through the axis at two distinct points — the equation has two real solutions. It can just touch the axis at exactly one point, where the lowest (or highest) point of the curve happens to land right on the axis — the equation has one solution, sometimes called a "repeated root." Or the parabola can sit entirely above the axis (or entirely below) without ever touching it — the equation has no real solutions at all. (It still has solutions, but they live in the complex numbers, which we'll come back to at the end.)

Which case happens depends on the specific values of a, b, c. There is a single number — called the discriminant — that decides. We will give it its proper name in the going-deeper section. For now, just hold on to the picture: three possibilities, controlled by where the parabola sits relative to the axis.

Solving the easy ones: factoring

Some quadratics let you read off the answer just by looking. Take

Ask yourself: are there two numbers that multiply to 6 and add to -5? Try a few. (-2) \times (-3) = 6 and (-2) + (-3) = -5. Got them. So you can rewrite the left side as

A product of two things is zero only if at least one of the things is zero. So either x - 2 = 0 or x - 3 = 0. The two solutions are x = 2 and x = 3. Done — and you didn't need a formula.

This works whenever the answers happen to be small integers and the middle coefficient is the right sum of the right factors. For about half the quadratics you will meet in early algebra, this is the fastest way. For the other half, the answers aren't pretty integers, and you need a tool that always works.

The trick that always works: completing the square

Here is the most beautiful idea in elementary algebra, and it is more than a thousand years old.

Suppose you have x^2 + 6x and you want to turn it into the square of something. You can't just write x^2 + 6x = (\text{something})^2 because (x + 3)^2 = x^2 + 6x + 9, which is not the same. There is a missing 9.

But you can add a 9. And the way to see why the missing piece is exactly 9 — and not, say, 18 — is to draw the algebra.

The geometric picture is doing real work: it shows you that the missing corner has to be exactly (b/2) \times (b/2), no more and no less. You can't argue with a square that fits. The deeper point is that some algebra is geometry in disguise — sometimes you can see a formula, not just compute it.

Now use this to solve a real equation. Take x^2 + 6x - 16 = 0 — the problem from the start of this article. Add 16 to both sides:

Add 9 to both sides (completing the square on the left):

The left side is now a perfect square:

Take the square root of both sides — and remember, taking a square root of a positive number gives you a \pm:

So x = 2 or x = -8 — exactly the two answers you found by guessing at the start of the article, now derived in a few clean lines. Completing the square never asks you to guess.

Completing the square is a recipe that always works on any quadratic. You can apply the same five steps mechanically to ax^2 + bx + c = 0 for any a, b, c. The result, in fully general form, is the formula every student eventually meets.

The quadratic formula

Apply completing the square to the general equation ax^2 + bx + c = 0. Five steps, exactly the same as before, just with letters instead of numbers.

Step 1. Divide both sides by a.

Why: the perfect-square pattern we are about to use only works cleanly when the coefficient of x^2 is 1. Dividing by a achieves that without changing what we are solving — both sides scale by the same factor, so the solutions don't move. (We can do this safely because a \neq 0; otherwise the equation isn't quadratic in the first place.)

Step 2. Move the constant term to the right side.

Why: completing the square is going to add something specific to the left side. We want only the x-terms there before that addition, so the constant gets moved out of the way first.

Step 3. Complete the square on the left.

Why: the same rule as in the geometric proof above. The missing corner of the square is (\text{half the }x\text{-coefficient})^2. Half of b/a is b/(2a), and squaring that gives b^2/(4a^2). We add it to both sides to keep the equation balanced — what we add on the left we must add on the right.

Step 4. Recognise the left side as a perfect square.

Why: the pattern x^2 + 2px + p^2 = (x + p)^2 matches what we have on the left, with p = b/(2a). So the left side collapses into (x + b/(2a))^2. The right side combines over the common denominator 4a^2: writing -c/a as -4ac/(4a^2) and adding it to b^2/(4a^2) gives (b^2 - 4ac)/(4a^2). The numerator b^2 - 4ac — the thing under the square root in the formula — appears for the first time here, and it is going to be important.

Step 5. Take the square root of both sides and then solve for x.

Why: square-rooting both sides always allows both signs, and that \pm is the entire reason a quadratic equation has two solutions instead of one. The square root of 4a^2 in the denominator is just 2a, which combines with \sqrt{b^2 - 4ac} in the numerator to give the clean form. After that, subtracting b/(2a) from both sides isolates x, and combining the two fractions over the common denominator 2a gives the formula in its standard, single-fraction shape.

What you have just derived is the quadratic formula:

This is one of the most-printed equations in human history, and Indian mathematicians had a recognisable form of it by the 7th century — Brahmagupta wrote it out in a slightly different but equivalent way in the Brahmasphutasiddhanta. It is the complete solution to a problem that took two thousand years to fully understand.

The quantity under the square root, b^2 - 4ac, is called the discriminant, and it is the single number that decides which of the three cases from earlier you are in. If it is positive, \sqrt{\,\cdot\,} is a real number and you get two distinct real solutions. If it is zero, the \pm becomes \pm 0 — there is only one solution, x = -b/(2a). If it is negative, the square root is not a real number and there are no real solutions. The three cases of the parabola, exactly recovered.

Computing one from start to finish

Time to do one yourself. We will do two examples — one that factors cleanly, and one that doesn't.

Common confusions

A few traps that catch almost everyone the first time.

• "x^2 = 25 means x = 5." It also means x = -5. Both squares are 25. Whenever you take a square root to solve for x, you have to allow both signs. The \pm in the quadratic formula is exactly this rule, applied once and for all.

• "If the discriminant is negative, there is no answer." There is no real answer. There are still two solutions, and they live in the complex numbers (you'll see them as x = a \pm bi where i is the imaginary unit). Whether you call them "answers" depends on whether you have met complex numbers yet — by class 11 in the Indian curriculum, you definitely will.

• "Every quadratic looks like x^2 + bx + c = 0." Not quite — the leading coefficient a is not always 1. For example, 3x^2 - 5x + 1 = 0 is a perfectly valid quadratic with a = 3. The formula handles this case fine; just remember to use a when you compute the discriminant (b^2 - 4ac, not just b^2 - 4c) and when you divide by 2a.

• "To solve x^2 + 5x = 0, divide both sides by x." Tempting, and wrong. If you divide by x, you are silently assuming x \neq 0 — and you lose the solution x = 0, which is one of the two real answers (the other being x = -5). When you have a quadratic with no constant term, factor it instead: x(x + 5) = 0 gives x = 0 or x = -5. Two answers, both kept.

Going deeper

The discriminant, named

The number b^2 - 4ac has a name: the discriminant, often written D or \Delta (the Greek letter delta). It "discriminates" between the three cases:

Discriminant	Roots	The parabola
D > 0	Two distinct real roots	Cuts the axis at two different points
D = 0	One repeated real root	Just touches the axis at the vertex
D < 0	Two complex conjugate roots	Sits entirely above or below the axis

The discriminant is the single number that controls which world you are in. It is worth computing first, before any other algebra, because it tells you what kind of answer to expect — and it lets you spot mistakes early. If you compute D = 12 but then your final answer comes out as a clean integer, something is wrong (clean integers come from D being a perfect square).

Sum and product of roots

There is a remarkably useful fact about quadratics. If the two roots are \alpha and \beta, then

You can read the sum and the product of the roots straight off the coefficients, without solving the equation at all. These are called Vieta's formulas, after the 16th-century French mathematician François Viète who first wrote them down systematically.

Why do they hold? Start with the roots and build the equation:

Compare this with x^2 + (b/a)x + (c/a) = 0 — what you get by dividing the original equation through by a. The coefficients have to match, which gives \alpha + \beta = -b/a and \alpha\beta = c/a. That's the entire proof.

These formulas are useful in three ways. First, as a check on your answer — once you have the two roots, verify that they sum to -b/a. Second, for extracting information without solving — if a problem asks "what is \alpha^2 + \beta^2?", you can compute it as (\alpha + \beta)^2 - 2\alpha\beta from the coefficients alone. Third, they generalise: a cubic has three roots, and you can read off the sum, the sum of pairwise products, and the product, all from the coefficients. The pattern keeps going up the degrees.

When the discriminant is negative

If D < 0, then \sqrt{D} is not a real number — but it is a perfectly good complex number. The two solutions become

where i is the imaginary unit, the number whose square is -1. The two solutions form a complex conjugate pair: same real part, opposite imaginary parts. You will meet them properly in the article on complex numbers.

What happens beyond degree two?

Quadratic equations are the simplest equations with more than one solution. The next case up is the cubic — degree three, like ax^3 + bx^2 + cx + d = 0. There is a closed-form formula for cubics too, much messier than the quadratic formula, but it exists. There is even one for degree four.

And then something extraordinary happens at degree five. It is provably the case — not just an unsolved problem, but proved impossible — that there is no closed-form formula for the general fifth-degree equation using only +, -, \times, \div, and roots. The four classical operations and roots are simply not enough.

It is one of the most surprising results in mathematics. The quadratic formula is not just a clever trick — it sits inside a hierarchy that ends, suddenly and forever, at degree five. We mention it here because it puts the formula you just learned in its proper place: it is the simplest case of one of the deepest questions in algebra.

Where this leads next

The quadratic formula opens up a lot of doors. The most direct continuations:

• Discriminant and Nature of Roots — a closer look at b^2 - 4ac and what each case means geometrically, with worked examples of all three.
• Sum and Product of Roots — Vieta's formulas worked out properly, with applications to symmetric functions of the roots.
• Quadratic Functions and Parabolas — the same equation seen as a function, with vertex form, axis of symmetry, and minimum/maximum values.
• Quadratic Inequalities — when you ask not "where is this zero" but "where is this positive or negative."
• Cubic Equations — what happens at degree three, including a proper look at Cardano's formula.