Solving quadratic equations

Every quadratic equation you will meet in Class 10 boils down to one question: for which values of $x$ does $ax^2+bx+c=0$ become true? Sometimes the answer falls out in seconds because the expression splits neatly into two brackets; sometimes it refuses to split at all, and you need a formula that works on every quadratic without exception. Both routes are really the same idea approached from two directions — factorisation guesses the roots as neat bracket factors, while the quadratic formula derives them symbolically for the general case, so whenever a quadratic does factorise, the two methods land on identical answers. Word problems add a third skill: turning a sentence about ages, consecutive numbers, or a rectangle's area into an equation, solving it, and then filtering the algebra through common sense — a negative age or a fractional number of people is a sign to reject that root, not to report it. This article builds the standard form, derives the quadratic formula from first principles, and works through both a clean factorisation and a word problem that needs the formula, before rounding up the mistakes that cost the most marks.

The core insight

Every quadratic $ax^2+bx+c=0$ can be solved either by splitting the middle term into two brackets (when it factorises neatly) or by the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ , which always works and reduces to the same roots as factorisation whenever both methods apply.

The core idea

A quadratic equation in one variable, once every term is collected on one side, always reduces to the standard form

ax^2+bx+c=0,\qquad a\ne 0.

where: $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , and $c$ is the constant term. The condition $a\ne 0$ is what makes it quadratic rather than linear — if $a=0$ the $x^2$ term vanishes and you are left with a straight-line equation instead.

Factorisation (Method 1). Rewrite the left side as a product of two linear brackets equal to zero, then invoke the zero-product rule: if $PQ=0$ then either $P=0$ or $Q=0$ . In practice this means splitting the middle term $bx$ into two parts whose coefficients multiply to give $ac$ and add to give $b$ , then grouping in pairs. This route is fast, but only when the quadratic happens to split into rational factors — not every quadratic does.

Quadratic formula (Method 2). This works on every quadratic, factorisable or not:

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.

where: $b^2-4ac$ is the discriminant, written $D$ . It must be zero or positive for real roots to exist, since a negative value under the square root has no real square root. A quick glance at $D$ before you start is worth the ten seconds it takes: if $D$ is a perfect square, factorisation will almost certainly go quickly; if it isn't, go straight to the formula rather than hunting for integer factors that don't exist.

Word problems. Translate the given conditions into an equation in one unknown, simplify to standard form, solve by whichever method fits, and check both roots against the real-world constraints of the question — reject anything that gives a negative length, a negative time, or a fractional count of objects or people.

Figure

A flowchart showing a quadratic equation branching into factorisation when the discriminant is a perfect square and the quadratic formula otherwise, both arrows meeting at the same pair of roots

Derivation ADeriving the quadratic formula by completing the square

The goal is to show that the formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ is not a rule to memorise blindly but a direct consequence of completing the square on the general equation $ax^2+bx+c=0$ , so that it is guaranteed to work on any quadratic, including the ones that stubbornly refuse to factorise into neat integer brackets.

Step 1. Start from $ax^2+bx+c=0$ and divide every term by $a$ (allowed since $a\ne 0$ ), giving $x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$ . Move the constant term to the right: $x^2+\dfrac{b}{a}x=-\dfrac{c}{a}$ .

Step 2. Complete the square on the left. Take half the coefficient of $x$ , which is $\dfrac{b}{2a}$ , square it to get $\dfrac{b^2}{4a^2}$ , and add this to both sides: $x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{b^2}{4a^2}-\dfrac{c}{a}$ . The left side is now a perfect square: $\left(x+\dfrac{b}{2a}\right)^2$ .

Step 3. Simplify the right side over a common denominator $4a^2$ : $\dfrac{b^2}{4a^2}-\dfrac{c}{a}=\dfrac{b^2-4ac}{4a^2}$ . So $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}$ . Take the square root of both sides, remembering both signs: $x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ .

Step 4. Isolate $x$ by subtracting $\dfrac{b}{2a}$ from both sides: $x=-\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . This is exactly the quadratic formula, and because every step used only algebra valid for any real $a\ne 0,b,c$ , it holds for every quadratic — factorisable or not.

Worked example

Question: Solve $x^2-5x-24=0$ by factorisation, then use the same equation to confirm the quadratic formula gives identical roots.

Step 1. Find two numbers whose product is $-24$ (the constant term, since the coefficient of $x^2$ is $1$ ) and whose sum is $-5$ (the coefficient of $x$ ). Those numbers are $-8$ and $3$ , since $(-8)(3)=-24$ and $-8+3=-5$ .

Step 2. Split the middle term using these numbers: $x^2-8x+3x-24=0$ . Group in pairs and factor each pair: $x(x-8)+3(x-8)=0$ , which gives $(x-8)(x+3)=0$ .

Step 3. Apply the zero-product rule: $x-8=0$ or $x+3=0$ , so $x=8$ or $x=-3$ . Check: $8^2-5(8)-24=64-40-24=0$ , and $(-3)^2-5(-3)-24=9+15-24=0$ . Both roots satisfy the equation.

Step 4. Confirm with the quadratic formula, using $a=1,\ b=-5,\ c=-24$ : $x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(-24)}}{2(1)}=\dfrac{5\pm\sqrt{25+96}}{2}=\dfrac{5\pm\sqrt{121}}{2}=\dfrac{5\pm 11}{2}$ . So $x=\dfrac{16}{2}=8$ or $x=\dfrac{-6}{2}=-3$ — the same roots, $x=8$ and $x=-3$ , exactly as factorisation gave, confirming the two methods agree whenever a quadratic factorises over rationals.

Competitive stretch

The product of two consecutive positive integers is $132$ . Let the smaller integer be $x$ , so $x(x+1)=132$ , giving $x^2+x-132=0$ . Here $a=1,b=1,c=-132$ , and the numbers needed to factorise aren't obvious at a glance, so use the formula: $x=\dfrac{-1\pm\sqrt{1+528}}{2}=\dfrac{-1\pm\sqrt{529}}{2}=\dfrac{-1\pm 23}{2}$ , giving $x=11$ or $x=-12$ . Since the question asks for positive integers, reject $x=-12$ (even though it is a mathematically valid root, since $(-12)(-11)=132$ too). So the integers are $\boxed{11 \text{ and } 12}$ , and $11\times 12=132$ checks out.

Common traps

Not rearranging into standard form $ax^2+bx+c=0$ (everything on one side, zero on the other) before factorising or substituting into the formula.
Sign errors when substituting negative $b$ or $c$ into the quadratic formula — always substitute in brackets first, as in $-(-5)=5$ above.
Accepting every root a word problem produces without checking it against the real-world context — a negative length, negative age, or non-whole count must be rejected even if it solves the equation correctly.
Grinding away at factorisation by trial and error when the discriminant is not a perfect square — check $b^2-4ac$ early and switch to the formula if it isn't a perfect square.

Ask the tutor: Ask the AI tutor to walk through completing the square on your own equation, or to help you decide quickly whether factorisation or the formula fits a specific word problem.

Why do factorisation and the quadratic formula always give the same answer

The quadratic formula is derived by completing the square on the general equation $ax^2+bx+c=0$ , so it is really factorisation carried out symbolically for every possible $a$ , $b$ , $c$ at once. Whenever a quadratic happens to split into rational-number brackets, both methods are describing the same two roots — one by spotting the split-numbers directly, the other by working through the algebra mechanically. If the roots are irrational or involve fractions that are hard to spot by inspection, factorisation becomes impractical and only the formula is convenient, but both are always mathematically consistent.

How do I know whether to try factorisation first or jump straight to the formula

Compute the discriminant $b^2-4ac$ first. If it is a perfect square (like $121$ or $529$ in the worked examples above), the quadratic will factorise into rational brackets and splitting the middle term is usually the faster route. If it isn't a perfect square, don't waste time guessing factor pairs — go straight to the quadratic formula, which works regardless. With practice, recognising perfect squares becomes quick, but when in doubt the formula is always a safe fallback.

In a word problem, why does the equation sometimes give two valid-looking roots but only one is the answer

A quadratic equation is purely algebraic and doesn't know anything about the real-world story behind it, so it will happily return both roots that satisfy the equation even if only one makes sense in context. For example, in the consecutive-integers example, $x=-12$ solves $x^2+x-132=0$ just as validly as $x=11$ does, because $-12$ and $-11$ are also consecutive integers whose product is $132$ . The question asked for positive integers, so the negative root has to be discarded as the final step — checking context against both roots is graded separately from the algebra and is a common place to lose marks even with fully correct working.

Solving quadratic equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads