What is the standard form of a quadratic equation?

ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. Any equation must be fully expanded and simplified before checking whether it truly reduces to this form — some equations that look quadratic actually simplify to linear, and vice versa.

What is the quadratic formula?

For ax² + bx + c = 0, the roots are x = [−b ± √(b² − 4ac)] / 2a, provided b² − 4ac ≥ 0. This formula works for any quadratic equation, unlike factorisation which only works when the equation splits cleanly.

What is the discriminant of a quadratic equation?

The discriminant is b² − 4ac. Its sign predicts the nature of the roots without solving the equation: positive means two distinct real roots, zero means two equal real roots, and negative means no real roots.

How do I solve a quadratic equation by factorisation?

Split the middle term (bx) into two parts whose product equals a×c and whose sum equals b, then factor by grouping into two linear factors. Setting each factor to zero gives the roots.

Why do I sometimes need to reject one root of a quadratic equation?

The quadratic formula returns mathematically valid roots regardless of context. In real-world word problems (lengths, ages, distances, counts), a negative root is often physically impossible and must be discarded, keeping only the root that makes sense.

Chapter 4 · CBSE Class 10 Maths

Quadratic Equations

Two ways to learn — a crisp formula reference, or a story that makes the discriminant feel obvious.

FactorisationQuadratic formulaDiscriminantNature of roots

Also useful:Chapter 3 — Pair of Linear Equations Chapter 5 — Arithmetic Progressions NCERT Solutions — Class 10 Maths 1-on-1 CBSE Class 10 Tuition

CBSEClass 10MathematicsChapter 4⏱ 8 min crisp · 12 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: quadratic equations show up disguised in real problems, and two tools — factorisation and the quadratic formula — always crack them, while the discriminant tells you what to expect before you even solve. Read the story, then tap "The Maths" for the formal result.

The Trickster Equation

Two equations, side by side. First: x(x+1) + 8 = (x+2)(x−2). It has x² terms glaring out at you from both sides — surely quadratic? Second: (x+2)³ = x³ − 4. This one has an x³ — surely NOT quadratic, right?

Teacher

Don't judge an equation by its appearance. Expand both sides fully before deciding anything.

Eqn 1: x²+x+8 = x²−4 → the x² terms CANCEL → x+12 = 0 (LINEAR, not quadratic!)

Eqn 2: x³+6x²+12x+8 = x³−4 → the x³ terms CANCEL → 6x²+12x+12=0 (QUADRATIC after all!)

Both equations were disguised — one hid a quadratic-turned-linear surprise, the other hid a cubic-turned-quadratic one. The lesson: the "degree" of an equation is only decided AFTER you've expanded and cancelled everything you can.

Breaking It Apart

A charity wants to build a prayer hall with area 300 m², where the length is one metre more than twice the breadth. Call the breadth x — then length = 2x+1, and area = x(2x+1) = 300, giving 2x² + x − 300 = 0.

Student

That's a real quadratic now. How do I actually find x from here?

2x² + x − 300 = 0

Split the middle term: 2x² − 24x + 25x − 300 = 0

2x(x − 12) + 25(x − 12) = 0 → (x − 12)(2x + 25) = 0

x = 12 or x = −12.5

Two mathematically valid answers — but x is a breadth, and breadth can't be negative. Reject x = −12.5 on physical grounds, not algebraic ones.

This is the same idea from Chapter 2: the roots of ax²+bx+c=0 are exactly the zeroes of the quadratic polynomial ax²+bx+c. Factorising to solve an equation is the identical move as factorising to find a polynomial's zeroes.

The Universal Key

A pole must be placed on the boundary of a circular park (diameter 13 m) so that its distances from two diametrically opposite gates differ by 7 m. Setting up the geometry (using the fact that any angle in a semicircle is 90°) gives x² + 7x − 60 = 0, where x is the distance to one gate.

Student

This one doesn't factorise as obviously. Is there a method that always works, no matter how ugly the numbers are?

x² + 7x − 60 = 0 → a=1, b=7, c=−60

Discriminant: b²−4ac = 49 − 4(1)(−60) = 49 + 240 = 289

x = (−7 ± √289) / 2 = (−7 ± 17) / 2

Two roots fall out: x = 5 or x = −12. Since x is a physical distance, reject the negative one — the pole sits 5 m from one gate and 12 m from the other. Quick sanity check using the right angle at the pole: 5² + 12² = 25 + 144 = 169 = 13², exactly matching the 13 m diameter used as the hypotenuse.

Reading the Discriminant

Before even reaching for the full quadratic formula, the expression under the square root — b²−4ac — already tells you what kind of answer to expect. Take 2x² − 4x + 3 = 0.

Teacher

Compute just the discriminant first, before solving anything. What does its sign tell you?

a=2, b=−4, c=3

D = (−4)² − 4(2)(3) = 16 − 24 = −8

D is NEGATIVE

A negative number has no real square root — so the ± √D part of the quadratic formula is asking for something that doesn't exist. This equation has NO real roots at all; don't waste time trying to factorise it.

Now compare with 3x² − 2x + 1/3 = 0, where D turns out to be exactly 0. Since √0 = 0, the ± vanishes entirely — both roots become the SAME value, −b/2a.

D = (−2)² − 4(3)(1/3) = 4 − 4 = 0

x = −(−2)/(2×3) = 2/6 = 1/3 (a single, repeated root)

D=0 is NOT the same as "no solution." It means exactly one value works, doubly. Only D<0 means no real solution exists at all.

The Big Picture

Every idea in this chapter branches from one habit: simplify fully before classifying, then pick a solving tool. Factorisation is fast when it works; the quadratic formula always works; the discriminant tells you the outcome before you even solve.

Is this equation really quadratic?

Expand fully first, then check the highest power

Fastest way to solve, if possible?

Factorisation: split into (px+q)(rx+s)=0

Method that always works?

x = [−b ± √(b²−4ac)] / 2a

Predict root count without solving?

Check the sign of D = b²−4ac

Got a negative root in a word problem?

Reject it if it violates a real-world constraint

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