CBSE Class 10 Maths — Quadratic Equations

Chapter Overview & Weightage

Quadratic Equations is one of the most heavily tested chapters in CBSE Class 10 Maths. It bridges algebra and real-world applications — from calculating dimensions to finding time and distance. Mastering this chapter is also essential preparation for Class 11 (where quadratics reappear in conic sections, sequences, and beyond).

In CBSE Class 10 board exams, Quadratic Equations typically carries 8–10 marks. The chapter appears across all question types: 1-mark MCQs (nature of roots), 2-mark questions (finding roots by factorisation), and 4-mark word problems (real-life applications). The discriminant and word problems are the two highest-mark areas.

Topics covered:

Standard form of a quadratic equation
Solving by factorisation
Solving by completing the square
Quadratic formula
Nature of roots (using discriminant)
Real-life word problems

Key Concepts You Must Know

1. Standard Form A quadratic equation in standard form is $ax^2 + bx + c = 0$ where $a \neq 0$ and $a, b, c$ are real numbers.

2. Methods of Solution Three methods: factorisation (fastest when it works), completing the square (always works, shows why the formula works), quadratic formula (always works, fastest for irrational roots).

3. Discriminant For $ax^2 + bx + c = 0$ , the discriminant is $D = b^2 - 4ac$ .

$D$ value	Nature of roots
$D > 0$	Two distinct real roots
$D = 0$	Two equal real roots (one root, repeated)
$D < 0$	No real roots (complex roots)

4. Relationship between roots (Vieta’s formulas — see Polynomials chapter): Sum of roots = $-b/a$ ; Product of roots = $c/a$ .

Important Formulas

For $ax^2 + bx + c = 0$ :

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

Use this when factorisation is not obvious or roots are irrational.

D = b^2 - 4ac

$D > 0$ : 2 distinct real roots | $D = 0$ : 2 equal roots | $D < 0$ : no real roots

Solved Previous Year Questions

PYQ 1 — Factorisation (CBSE 2023)

Solve: $2x^2 - 5x + 3 = 0$

Solution: $2x^2 - 5x + 3 = 0$

Find two numbers with product $= 2 \times 3 = 6$ and sum $= -5$ : these are $-2$ and $-3$ .

$2x^2 - 2x - 3x + 3 = 0$

$2x(x - 1) - 3(x - 1) = 0$

$(2x - 3)(x - 1) = 0$

$x = 3/2$ or $x = 1$

PYQ 2 — Discriminant and nature of roots (CBSE 2024 Style)

Find the value of $k$ for which $kx^2 + 2kx + 3 = 0$ has equal roots.

Solution: For equal roots: $D = 0$

$D = (2k)^2 - 4 \cdot k \cdot 3 = 4k^2 - 12k = 0$

$4k(k - 3) = 0$

$k = 0$ or $k = 3$

Since $k = 0$ makes it a linear (not quadratic) equation, we reject $k = 0$ .

$\boxed{k = 3}$

PYQ 3 — Word problem (CBSE 2023, 4 marks)

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the original speed.

Solution: Let original speed = $x$ km/h.

Time at original speed = $360/x$ hours.

Time at new speed = $360/(x+5)$ hours.

Given: $\frac{360}{x} - \frac{360}{x+5} = 1$

$360(x+5) - 360x = x(x+5)$

$360x + 1800 - 360x = x^2 + 5x$

$x^2 + 5x - 1800 = 0$

$(x + 45)(x - 40) = 0$

$x = -45$ (rejected, speed can’t be negative) or $x = 40$ km/h.

Original speed = 40 km/h

Difficulty Distribution

Question Type	Marks	Difficulty	Frequency
Find roots (factorisation)	2	Easy	Very High
Find roots (quadratic formula)	2	Easy–Medium	High
Nature of roots / discriminant	2	Easy	High
Find $k$ for equal/no real roots	2	Medium	High
Real-life word problem	4	Medium–Hard	High
Completing the square method	3	Medium	Low

Expert Strategy

Factorisation should always be tried first. For most CBSE problems, the coefficients are small integers and factorisation works in 30 seconds. Only switch to the formula if factorisation isn’t obvious within 1 minute.

For word problems, always define a variable, write the equation from the given condition, simplify to standard form, then solve. The most common word problem types are: number problems, age problems, speed-time-distance, and area/dimensions.

Discriminant questions are easy marks — the formula is simple and they always check $D > 0$ , $D = 0$ , or $D < 0$ . Don’t skip these in practice.

In word problems, always reject negative solutions if they represent physical quantities (distance, speed, age, number of items). For example, if you get $x = -5$ or $x = 3$ for a “number of students,” reject $-5$ and write “number of students = 3.” Examiners expect you to state the rejection explicitly.

Common Traps

Trap 1 — Dividing by $x$ to simplify. If you see $x^2 + 5x = 0$ and divide by $x$ , you get $x + 5 = 0 \implies x = -5$ . But you’ve lost the solution $x = 0$ ! Always factorise: $x(x+5) = 0 \implies x = 0$ or $x = -5$ . Never divide a quadratic by the variable.

Trap 2 — Completing the square sign errors. To complete the square for $x^2 + 6x - 7$ : $x^2 + 6x = (x+3)^2 - 9$ . A common error is forgetting to subtract the square term: writing $(x+3)^2 - 7$ instead of $(x+3)^2 - 9 - 7 = (x+3)^2 - 16$ . Always expand to verify.

Trap 3 — Wrong formula for discriminant. The discriminant is $b^2 - 4ac$ for $ax^2 + bx + c = 0$ . For $2x^2 - 3x + 1 = 0$ : $a = 2, b = -3, c = 1$ . $D = (-3)^2 - 4(2)(1) = 9 - 8 = 1$ . A common mistake is using $b = 3$ (forgetting the sign), giving $D = 9 - 8 = 1$ — same answer here, but not always.

Trap 4 — Word problem equation setup. The most common error is setting up the wrong equation. Read the problem twice. “Three years ago, John’s age was half his brother’s age” means $(x-3) = \frac{1}{2}(y-3)$ , not $x-3 = \frac{1}{2}y$ . The “three years ago” applies to BOTH ages.