§4Class 10, Chapter 4

Quadratic Equations: Forecasting the Roots

Every quadratic equation hides a built-in forecast of its own answers — the discriminant tells you how many real roots to expect before you even solve for them. This companion connects factorisation to the quadratic formula, and gives you tools, flashcards and a CBSE-style quiz to test yourself.

ax²+bx+c=0b²−4acαβ

🏗️4.1 From polynomials to equations: a prayer hall problem

A quadratic polynomial ax² + bx + c (with a ≠ 0) is just an expression — but the moment you ask "for which x does this expression equal zero?", you've turned it into a quadratic equation: ax² + bx + c = 0.

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A polynomial asking a question. Think of a polynomial as a statement ("here's a formula") and an equation as a question ("for what input does this formula give zero?"). The zeroes of the polynomial from Chapter 2 and the roots of the equation in this chapter are literally the same numbers — this whole chapter is about finding them.

Here's where it shows up in real life: a charity wants to build a prayer hall with carpet area 300 m², where the length is one metre more than twice the breadth. If the breadth is x metres, the length is (2x+1) metres, so:

area = (2x + 1) × x = 2x² + x

2x² + x = 300

2x² + x − 300 = 0

So the breadth x must satisfy this quadratic equation. Finding x — and therefore the hall's actual dimensions — is exactly the kind of problem this chapter solves (we'll finish this one later, in Section 4.3).

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A very old problem. Quadratic equations are ancient. Babylonians could already find two numbers with a given sum and product (equivalent to solving x² − px + q = 0). Brahmagupta (c. 598–665 CE) gave an explicit formula for ax² + bx = c. Sridharacharya (c. 1025 CE) derived what we now call the quadratic formula, by a method called "completing the square." This isn't a modern invention — it's a problem people have been solving for well over a thousand years.

🧮4.2 What actually makes an equation quadratic

Standard form. A quadratic equation in x is any equation that can be written as ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. Writing the terms in descending powers of x gives the standard form.

⚠️Looks can be deceiving. An equation can look quadratic (or cubic!) before you simplify it, and turn out to be something else entirely once you actually expand and collect terms. Always simplify fully before deciding what kind of equation you're dealing with.

🎭Example: costumes that don't match what's underneath

Check whether x(x+1) + 8 = (x+2)(x−2) is a quadratic equation:

LHS = x(x+1) + 8 = x² + x + 8

RHS = (x+2)(x−2) = x² − 4

So: x² + x + 8 = x² − 4 → x + 12 = 0

The x² terms on both sides cancel out completely, leaving x + 12 = 0 — a linear equation, not quadratic, even though it started out looking like one.

Now check (x+2)³ = x³ − 4:

LHS = (x+2)³ = x³ + 6x² + 12x + 8

So: x³ + 6x² + 12x + 8 = x³ − 4 → 6x² + 12x + 12 = 0 → x² + 2x + 2 = 0

Here the opposite happens: the x³ terms cancel, and what looked like a cubic equation turns out to be quadratic after all.

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Simplify before you judge. It's like a costume party — some guests dressed as something they're not. An equation's degree is only revealed once every duplicate power of x on both sides has been cancelled out. Never classify an equation from its unsimplified appearance.

📝Translating word problems into equations

John and Jivanti together have 45 marbles. Each loses 5, and the product of what's left is 124. Letting x = John's original marbles, Jivanti had (45 − x):

(x − 5)(40 − x) = 124

−x² + 45x − 200 = 124

x² − 45x + 324 = 0

A cottage industry's daily toy cost per toy is (55 − number produced); the total cost one day was ₹750. Letting x = toys produced:

x(55 − x) = 750

x² − 55x + 750 = 0

Both situations reduce to a quadratic equation — this translation step (words → algebra) is often the hardest part of solving these problems, harder than the algebra that follows.

✂️4.3 Solving by factorisation

Root of a quadratic equation. A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. This is exactly the same condition as α being a zero of the polynomial ax² + bx + c — roots of the equation and zeroes of the polynomial are the same numbers. Since a quadratic polynomial has at most 2 zeroes (Chapter 2), a quadratic equation has at most 2 roots.

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Factorising is "un-multiplying". If ax² + bx + c can be split into two linear factors, like (px + q)(rx + s), then the product is zero exactly when either factor is zero — because the only way to multiply two things and get zero is if one of them already was zero. That single idea turns one quadratic equation into two much easier linear ones.

Worked example: 2x² − 5x + 3 = 0

SPLITSplit the middle term −5x into −2x and −3x, chosen so their product matches (2x²)×3 = 6x²: −5x = −2x − 3x.

GROUP2x² − 2x − 3x + 3 = 2x(x − 1) − 3(x − 1) = (2x − 3)(x − 1).

SOLVE EACH FACTOR(2x − 3)(x − 1) = 0 means 2x − 3 = 0 or x − 1 = 0, giving x = 3/2 or x = 1.

Worked example: 6x² − x − 2 = 0

6x² − x − 2 = 6x² + 3x − 4x − 2 = 3x(2x+1) − 2(2x+1) = (3x−2)(2x+1)

Roots: x = 2/3 or x = −1/2

Worked example: 3x² − 2√6x + 2 = 0 (a repeated root)

3x² − 2√6x + 2 = 3x² − √6x − √6x + 2 = √3x(√3x − √2) − √2(√3x − √2) = (√3x − √2)(√3x − √2)

Both factors are identical, so there's only one distinct value: √3x − √2 = 0 gives x = √(2/3). This root is repeated — it counts twice, once for each copy of the factor, even though only one distinct number shows up.

🏗️Back to the prayer hall

Recall from Section 4.1: 2x² + x − 300 = 0, where x is the hall's breadth.

2x² − 24x + 25x − 300 = 0

2x(x − 12) + 25(x − 12) = 0

(x − 12)(2x + 25) = 0

→ x = 12 or x = −12.5

📏Discarding the impossible root. x = −12.5 is a perfectly valid algebraic solution — but x represents a physical breadth, and a negative breadth makes no sense. So we discard it, keeping only x = 12. The hall's breadth is 12 m and its length is 2(12)+1 = 25 m. Algebra doesn't know about real-world constraints — checking and discarding invalid roots is a step you have to do yourself.

🔍4.4 Nature of roots: the discriminant

Factorising works well when the numbers are friendly, but not every quadratic factors neatly. The quadratic formula solves any quadratic equation directly from its coefficients:

x = [ −b ± √(b² − 4ac) ] / 2a

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A weather forecast for your roots. The expression under the square root, b² − 4ac, is called the discriminant. You can check its sign before doing any more work, and it tells you in advance exactly what kind of answer you're going to get — like checking the forecast before deciding whether to bring an umbrella.

Three cases. (i) If b² − 4ac > 0 → the equation has two distinct real roots. (ii) If b² − 4ac = 0 → the equation has two equal (repeated) real roots, both equal to −b/2a. (iii) If b² − 4ac < 0 → the equation has no real roots (you'd need to take the square root of a negative number).

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This is the same story as Chapter 2's parabolas. These three cases are exactly the three parabola shapes from Chapter 2: a positive discriminant means the parabola crosses the x-axis twice (two zeroes); a zero discriminant means it just touches the x-axis once (one repeated zero); a negative discriminant means it never reaches the x-axis at all (no real zero). The discriminant is just a numerical shortcut for reading the graph without drawing it.

b²−4ac > 0 — two distinct roots

b²−4ac = 0 — two equal roots

b²−4ac < 0 — no real roots

Worked example: 2x² − 4x + 3 = 0

a=2, b=−4, c=3

b² − 4ac = 16 − 24 = −8 < 0

→ No real roots.

🎡Real-world example: the pole and the two gates

A circular park has diameter 13 m, with gates A and B at opposite ends. A pole must go on the boundary so its distances from A and B differ by 7 m. Letting x = distance from B, and using the fact that ∠APB = 90° (angle in a semicircle) with the Pythagorean theorem:

SET UPAP = (x+7), BP = x, AB = 13. By Pythagoras: (x+7)² + x² = 13² → 2x² + 14x − 120 = 0 → x² + 7x − 60 = 0.

CHECK DISCRIMINANTb² − 4ac = 7² − 4(1)(−60) = 49 + 240 = 289 > 0 → real roots exist, so the pole placement is possible.

APPLY FORMULAx = (−7 ± √289)/2 = (−7 ± 17)/2 → x = 5 or x = −12.

DISCARD & CONCLUDEx is a distance, so x = −12 is rejected. x = 5. The pole sits 5 m from gate B and 12 m from gate A.

Worked example: 3x² − 2x + 1/3 = 0 (equal roots)

a=3, b=−2, c=1/3

b² − 4ac = 4 − 4(3)(1/3) = 4 − 4 = 0 → two equal real roots

Both roots = −b/2a = 2/6 = 1/3

📋4.5 Summary

🧮 A quadratic equation in x has the form ax² + bx + c = 0, with a, b, c real and a ≠ 0.
🔗 A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. Roots of the equation and zeroes of the corresponding polynomial are the same numbers.
✂️ If ax² + bx + c factorises into two linear factors, the roots come from setting each factor to zero.
📐 Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a, valid whenever b² − 4ac ≥ 0.
🔍 A quadratic equation has two distinct real roots if b² − 4ac > 0, two equal real roots if b² − 4ac = 0, and no real roots if b² − 4ac < 0.