Class 10 Maths Quadratic Equations: Important Questions, Competency Problems & Solutions

Quadratic Equations questions come in two genuinely different flavours, and most practice sets only give you one of them. There's the traditional skill — factorize this, find the discriminant of that, solve this word problem — and there's the newer competency-based format, where a single real-world scenario asks you to model a situation as a quadratic equation before you even start solving. Since 50% of CBSE Class 10 papers are now competency-based, both deserve real practice, not just the first one.

How to use this page: Attempt each problem fully on paper before tapping "Reveal Solution" — for the case studies, work through all sub-questions before checking any of them. Every calculation on this page was computed and independently verified before publishing.

Part A — Important Questions

8 core question types, 4 problems each — the traditional skills every quadratic equations paper tests.

4 Case Studies

Part B — Competency Scenarios

12 linked sub-questions across 4 real-world case studies, matching CBSE's actual case-based format.

Part A: The 8 Core Question Types

Nature of roots (discriminant)

Solving by factorization

Solving by the quadratic formula

Finding an unknown constant from root conditions

Word problems — numbers

Word problems — ages

Word problems — speed, distance, time

Word problems — geometric / area

Type 1 · Nature of Roots

1.1

No Real Roots

Find the nature of the roots of 2x² − 4x + 3 = 0.

Reveal Solution

D = b²−4ac = (−4)² − 4(2)(3) = 16 − 24 = −8

✓ D < 0 → No real roots exist.

1.2

Equal Roots

Find the nature of the roots of 4x² − 4x + 1 = 0. If real and equal, find the root.

Reveal Solution

D = (−4)² − 4(4)(1) = 16 − 16 = 0

Since D = 0

Root = −b/2a = 4/8

✓ D = 0 → Real and equal roots. Root = 1/2.

1.3

Rational Roots

Find the nature of the roots of 2x² − 5x + 3 = 0.

Reveal Solution

D = (−5)² − 4(2)(3) = 25 − 24 = 1

✓ D > 0 and a perfect square → Real, distinct, rational roots (3/2 and 1).

1.4

Irrational Roots

Find the nature of the roots of 2x² − 6x + 3 = 0.

Reveal Solution

D = (−6)² − 4(2)(3) = 36 − 24 = 12

✓ D > 0 but not a perfect square → Real, distinct, irrational roots.

⚠️ Common mistake: stopping at "D > 0, so two real roots" without checking whether D is a perfect square — this distinction (rational vs irrational roots) is specifically what many questions ask for.

Type 2 · Factorization

2.1

Factorize

Solve by factorization: x² − 7x + 12 = 0

Reveal Solution

Find two numbers multiplying to 12, adding to −7

−3 and −4 work: (−3)×(−4)=12, (−3)+(−4)=−7

x² − 3x − 4x + 12 = 0 → x(x−3) − 4(x−3) = 0 → (x−3)(x−4) = 0

✓ x = 3 or x = 4

2.2

Factorize

Solve by factorization: 2x² + 5x − 3 = 0

Reveal Solution

Find two numbers multiplying to (2×−3)=−6, adding to 5

6 and −1 work

2x² + 6x − x − 3 = 0 → 2x(x+3) − 1(x+3) = 0 → (x+3)(2x−1) = 0

✓ x = −3 or x = 1/2

2.3

Factorize

Solve by factorization: 6x² − x − 1 = 0

Reveal Solution

Find two numbers multiplying to (6×−1)=−6, adding to −1

−3 and 2 work

                                6x² − 3x + 2x − 1 = 0 → 3x(2x−1) + 1(2x−1) = 0 → (2x−1)(3x+1) = 0
                            

✓ x = 1/2 or x = −1/3

2.4

Factorize

Solve by factorization: x² − 3x − 10 = 0

Reveal Solution

Find two numbers multiplying to −10, adding to −3

−5 and 2 work

(x−5)(x+2) = 0

✓ x = 5 or x = −2

Type 3 · Quadratic Formula

3.1

Formula Method

Solve using the quadratic formula: x² − 6x + 4 = 0

Reveal Solution

D = 36 − 16 = 20. x = (6 ± √20)/2

✓ x = 3 + √5 or x = 3 − √5

3.2

Formula Method

Solve using the quadratic formula: x² − 2x − 6 = 0

Reveal Solution

D = 4 + 24 = 28. x = (2 ± √28)/2 = (2 ± 2√7)/2

✓ x = 1 + √7 or x = 1 − √7

3.3

Formula Method

Solve using the quadratic formula: 2x² − 7x + 2 = 0

Reveal Solution

D = 49 − 16 = 33. x = (7 ± √33)/4

✓ x = (7+√33)/4 ≈ 3.19 or x = (7−√33)/4 ≈ 0.31

⚠️ Neither root is a "nice" number here, and that's fine — the quadratic formula exists precisely for equations like this one that don't factor cleanly.

3.4

Formula Method

Solve using the quadratic formula: x² + 4x − 4 = 0

Reveal Solution

D = 16 + 16 = 32. x = (−4 ± √32)/2 = (−4 ± 4√2)/2

✓ x = −2 + 2√2 or x = −2 − 2√2

Type 4 · Finding an Unknown Constant

4.1

Equal Roots Condition

Find the value of k for which kx² + 4x + 1 = 0 has equal roots.

Reveal Solution

Equal roots means D = 0

D = 4² − 4(k)(1) = 16 − 4k

16 − 4k = 0

✓ k = 4

4.2

Given Root Value

If x = 3 is a root of 2x² + kx + 3 = 0, find k.

Reveal Solution

Substitute x=3 into the equation

2(3)² + k(3) + 3 = 0 → 18 + 3k + 3 = 0

3k = −21

✓ k = −7

4.3

Sum-Product Condition

For x² − (k+6)x + 2(2k−1) = 0, find k if the sum of the roots is half the product of the roots.

Reveal Solution

Identify sum and product

Sum = k+6, Product = 2(2k−1)

k+6 = (1/2) × 2(2k−1) → k+6 = 2k−1

✓ k = 7

4.4

Sum-Product Condition

For 3x² + (2k+1)x + (k−5) = 0, find k if the product of the roots is twice the sum of the roots.

Reveal Solution

Identify sum and product

Sum = −(2k+1)/3, Product = (k−5)/3

(k−5)/3 = 2 × [−(2k+1)/3] → k−5 = −2(2k+1) → k−5 = −4k−2

✓ k = 3/5

Type 5 · Word Problems — Numbers

5.1

Consecutive Integers

The sum of the squares of two consecutive positive integers is 365. Find the integers.

Reveal Solution

Let the integers be x and x+1

x² + (x+1)² = 365

2x² + 2x − 364 = 0 → x² + x − 182 = 0 → (x−13)(x+14) = 0

                    ✓ x = 13 (rejecting x = −14 since integers are positive) → The integers are 13 and 14.

5.2

Consecutive Even Integers

The product of two consecutive positive even integers is 168. Find the integers.

Reveal Solution

Let the integers be x and x+2

x(x+2) = 168

x² + 2x − 168 = 0 → (x−12)(x+14) = 0

✓ x = 12 (rejecting negative root) → The integers are 12 and 14.

5.3

Two Numbers

One number is 5 more than another. If their product is 176, find the numbers.

Reveal Solution

Let the smaller number be x

x(x+5) = 176

x² + 5x − 176 = 0 → (x−11)(x+16) = 0

✓ x = 11 (rejecting negative root) → The numbers are 11 and 16.

5.4

Number & Reciprocal

The sum of a positive number and its reciprocal is 41/20. Find the number.

Reveal Solution

Let the number be x

x + 1/x = 41/20 → 20x² − 41x + 20 = 0

(4x−5)(5x−4) = 0

                    ✓ x = 5/4 or x = 4/5 (both valid — they're reciprocals of each other, so either is a correct "the number")

Type 6 · Word Problems — Ages

6.1

Age Product

The product of a person's age 5 years ago and their age 8 years from now is 30. Find their present age.

Reveal Solution

Let present age = x

(x−5)(x+8) = 30

x² + 3x − 40 − 30 = 0 → x² + 3x − 70 = 0 → (x−7)(x+10) = 0

✓ x = 7 (rejecting negative root) → Present age = 7 years.

6.2

Mother-Daughter

A mother's age is three times her daughter's age. The product of their ages is 300. Find the daughter's age.

Reveal Solution

Let daughter's age = x, mother's age = 3x

x(3x) = 300 → 3x² = 300 → x² = 100

                    ✓ x = 10 (rejecting negative root) → Daughter's age = 10 years, mother's age = 30 years.

6.3

Father-Son

The sum of a father's and son's present ages is 45 years. Five years ago, the product of their ages was 124. Find their present ages.

Reveal Solution

Let son's present age = x, so father's = 45−x

5 years ago: (x−5) and (40−x). Product = 124

                                (40−x)(x−5) = 124 → −x²+45x−324 = 0 → x²−45x+324 = 0 → (x−9)(x−36) = 0
                            

Both x=9 and x=36 solve the equation — but only one fits the context

If son's age = 36, father's age would be 9, which is impossible for a father. So reject x=36.

✓ Son's age = 9 years, father's age = 36 years.

⚠️ This is a genuinely important skill: both roots satisfy the equation mathematically, but only checking them against real-world logic (a father must be older than his son) tells you which one is the actual answer.

6.4

Woman-Son

Two years ago, a woman's age was three times the square of her son's age. Three years from now, her age will be four times her son's age. Find the son's present age.

Reveal Solution

Let son's present age = x

2 years ago, son's age = x−2, woman's age then = 3(x−2)². So woman's present age = 3(x−2)²+2.

3 years from now

Woman's age = 3(x−2)²+5, son's age = x+3. Condition: woman's age = 4 × son's age.

3(x−2)²+5 = 4(x+3) → 3x²−16x+5 = 0 → (3x−1)(x−5) = 0

Check both roots against context

x=1/3 would mean "2 years ago" the son's age was negative — impossible. Reject it.

✓ Son's present age = 5 years.

Type 7 · Word Problems — Speed, Distance, Time

7.1

Train Speed

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 for the journey. Find the original speed.

Reveal Solution

Let original speed = x km/h

360/x − 360/(x+5) = 1

                                360(x+5) − 360x = x(x+5) → 1800 = x²+5x → x²+5x−1800=0 → (x−40)(x+45)=0
                            

✓ x = 40 (rejecting negative root) → Original speed = 40 km/h.

7.2

Boat Speed

A boat covers 24 km upstream and 24 km downstream in a total of 5 hours. The speed of the stream is 2 km/h. Find the boat's speed in still water.

Reveal Solution

Let boat's speed in still water = x km/h

Upstream speed = x−2, downstream speed = x+2

                                24/(x−2) + 24/(x+2) = 5 → 24(x+2)+24(x−2) = 5(x−2)(x+2) → 48x = 5x²−20
                            

5x² − 48x − 20 = 0 → (5x+2)(x−10) = 0

✓ x = 10 (rejecting negative root) → Boat's speed in still water = 10 km/h.

7.3

Car Speed

A car covers a distance of 600 km. Had the speed been 10 km/h less, it would have taken 2 hours more for the same journey. Find the original speed.

Reveal Solution

Let original speed = x km/h

600/(x−10) − 600/x = 2

                                600x − 600(x−10) = 2x(x−10) → 6000 = 2x²−20x → x²−10x−3000=0 → (x−60)(x+50)=0
                            

✓ x = 60 (rejecting negative root) → Original speed = 60 km/h.

7.4

Pipes Filling a Tank

Two pipes together can fill a tank in 6 hours. If used separately, one pipe takes 5 hours more than the other. Find the time each pipe takes separately.

Reveal Solution

Let the faster pipe take x hours; the slower takes x+5 hours

1/x + 1/(x+5) = 1/6

                                6(x+5) + 6x = x(x+5) → 12x+30 = x²+5x → x²−7x−30=0 → (x−10)(x+3)=0
                            

                    ✓ x = 10 (rejecting negative root) → Faster pipe: 10 hours, Slower pipe: 15 hours.

Type 8 · Word Problems — Geometric / Area

8.1

Rectangle

The length of a rectangular field is 3 m more than its breadth. If the area is 154 sq. m, find the breadth.

Reveal Solution

Let breadth = x, length = x+3

x(x+3) = 154

x²+3x−154=0 → (x−11)(x+14)=0

✓ x = 11 (rejecting negative root) → Breadth = 11 m, Length = 14 m.

8.2

Right Triangle

The hypotenuse of a right triangle is 13 cm. One leg is 7 cm more than the other. Find the length of the shorter leg.

Reveal Solution

Let shorter leg = x, longer leg = x+7

x² + (x+7)² = 13² (Pythagoras)

2x²+14x+49=169 → 2x²+14x−120=0 → x²+7x−60=0 → (x−5)(x+12)=0

✓ x = 5 (rejecting negative root) → Shorter leg = 5 cm, longer leg = 12 cm.

8.3

Garden Dimensions

A rectangular garden has a perimeter of 44 m and an area of 112 sq. m. Find its dimensions.

Reveal Solution

From perimeter: length + breadth = 22

Let breadth = x, length = 22−x. Area: x(22−x) = 112

22x−x²=112 → x²−22x+112=0 → (x−8)(x−14)=0

                    ✓ Breadth = 8 m, Length = 14 m (the two roots simply correspond to which side you call "breadth" vs "length").

8.4

Two Squares

The sum of the perimeters of two squares is 64 cm, and the difference of their areas is 32 sq. cm. Find the side of the larger square.

Reveal Solution

From perimeter: sides sum to 16

Let larger side = x, smaller side = 16−x. Difference of areas: x² − (16−x)² = 32

x²−(256−32x+x²)=32 → 32x−256=32 → 32x=288

✓ x = 9 → Larger square's side = 9 cm, smaller square's side = 7 cm.

Part B · Competency-Based Case Studies

Each scenario below is a single real-world situation with three linked questions — exactly the format CBSE uses for competency-based, case-based questions. Work through all three sub-questions for a scenario before revealing any of the answers.

Case Study 1

The Bookshop

A shopkeeper buys a certain number of books for ₹80 in total. If he had bought 4 more books for the same ₹80, each book would have cost him ₹1 less than what he actually paid.

Sub-question (i)

Form the quadratic equation representing this situation, taking the original number of books as x.

Reveal Answer

Cost per book originally = 80/x. Cost per book with 4 more books = 80/(x+4). The difference is ₹1: 80/x − 80/(x+4) = 1

✓ Simplifying: x² + 4x − 320 = 0

Sub-question (ii)

Solve the equation to find the original number of books.

Reveal Answer

x²+4x−320=0 → (x−16)(x+20)=0

✓ x = 16 (rejecting the negative root) → 16 books originally.

Sub-question (iii)

Find the original cost per book.

Reveal Answer

Cost per book = 80/16

                            ✓ ₹5 per book (check: with 20 books, cost/book = 80/20 = ₹4, exactly ₹1 less ✓)

Case Study 2

The Bookshop

A ball is thrown upward from ground level. Its height h (in metres) after t seconds is modelled by h(t) = −5t² + 30t.

Sub-question (i)

After how many seconds does the ball return to the ground?

Reveal Answer

The ball is at ground level when h(t) = 0: −5t²+30t = 0 → −5t(t−6) = 0

                        ✓ t = 0 (the start) or t = 6. The ball returns to the ground at t = 6 seconds.

Sub-question (ii)

At what time(s) does the ball reach a height of 40 m?

Reveal Answer

−5t²+30t = 40 → −5t²+30t−40=0 → t²−6t+8=0 → (t−2)(t−4)=0

                        ✓ t = 2 seconds AND t = 4 seconds — both are valid, since the ball passes 40 m once on the way up and again on the way down.

Sub-question (iii)

Why does this problem have two valid answers for part (ii), while most word problems reject one root as impossible?

Reveal Answer

                        ✓ Because both moments are physically real — the ball genuinely passes through 40 m twice on its full path (once rising, once falling). This is different from problems like ages or lengths, where only one root fits reality; here, the situation itself has two real, meaningful events.

Case Study 3

The Bookshop

A municipal committee is designing a rectangular park. The length is planned to be 3 m more than twice its breadth. The total area available is 65 sq. m. The committee also needs to fence the park at a cost of ₹50 per metre.

Sub-question (i)

Form the quadratic equation for the breadth, b.

Reveal Answer

Length = 2b+3. Area = b(2b+3) = 65

✓ 2b² + 3b − 65 = 0

Sub-question (ii)

Find the breadth and length of the park.

Reveal Answer

2b²+3b−65=0 → (2b+13)(b−5)=0

                        ✓ b = 5 (rejecting the negative root) → Breadth = 5 m, Length = 2(5)+3 = 13 m.

Sub-question (iii)

Calculate the total fencing cost for the park.

Reveal Answer

Perimeter = 2(length+breadth) = 2(13+5) = 36 m

Cost = 36 × ₹50

✓ Total fencing cost = ₹1,800

Case Study 4

The Bookshop

A rectangular photograph measuring 16 cm by 8 cm is mounted on a rectangular cardboard sheet, leaving a uniform border of width x cm on all four sides. The total area of the cardboard sheet (photo plus border) is 240 sq. cm.

Sub-question (i)

Form the quadratic equation for the border width, x.

Reveal Answer

With the border, the total dimensions become (16+2x) by (8+2x), since the border adds x on both sides of each dimension.

✓ (16+2x)(8+2x) = 240

Sub-question (ii)

Solve for the border width.

Reveal Answer

                        128+32x+16x+4x²=240 → 4x²+48x−112=0 → x²+12x−28=0 → (x−2)(x+14)=0
                    

✓ x = 2 (rejecting the negative root) → Border width = 2 cm.

Sub-question (iii)

What are the outer dimensions of the cardboard sheet?

Reveal Answer

Outer length = 16+2(2) = 20 cm. Outer breadth = 8+2(2) = 12 cm.

✓ 20 cm × 12 cm (check: 20×12 = 240 sq. cm ✓)

The Skill Isn't Just Solving — It's Modelling and Interpreting Too

A student who can solve x²+4x−320=0 correctly might still struggle to set up that exact equation from the Bookshop scenario, or fail to reject the impossible root in an ages problem. These are three separate skills — modelling, solving, and interpreting — and a single overall score doesn't tell you which one is actually weak.

What a Genelis weak area map looks like after working through Quadratic Equations practice

Factorization & quadratic formula

85%

Word problems — solving once modelled

68%

Rejecting inadmissible roots

47%

Competency case studies — modelling from scratch

29%

Next session: competency case-study modelling (29%) — the exact skill a strong "solving" score can hide entirely. Genelis tracks these as separate skills, not one combined topic score.

Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh problems across both traditional question types and competency case studies, tracking modelling, solving, and interpretation as distinct skills. Every wrong answer is logged to your wrong-question notebook for reattempt.

Step 1

Attempt fresh problems

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Step 2

Skill-level gap detected

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Step 3

AI notes for weak pattern

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Step 4

Wrong Qs auto-logged

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Step 5

Reattempt that skill

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Result

Gap closed. Map updates. ✓

Practise unlimited fresh Quadratic Equations problems on Genelis — free →

💡 For chapter strategy and the complete competency-question format used across CBSE, see our Class 10 Competency-Based Questions guide.

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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What question types appear in CBSE Class 10 Quadratic Equations?

Eight traditional types recur most often: determining the nature of roots using the discriminant, solving by factorization, solving using the quadratic formula, finding an unknown constant from a condition on the roots, and four categories of word problems — numbers, ages, speed-distance-time, and geometric or area-based problems. Beyond these, CBSE also tests competency-based case study questions, which present a real-world scenario and ask you to first model it as a quadratic equation before solving it.

What is a competency-based question in Quadratic Equations, and how is it different from a word problem?

A competency-based question presents a single real-world scenario followed by two or three linked sub-questions, testing not just your ability to solve a quadratic equation but your ability to model a situation into one, interpret the result in context, and sometimes extend the answer to a further calculation. A traditional word problem usually asks for one final answer; a competency case study typically asks you to form the equation, solve it, and then apply the result — for example, using a park's dimensions to calculate a fencing cost.

Why do some quadratic word problems have two mathematical solutions but only one correct answer?

A quadratic equation always has two roots (or one repeated root), but not every root makes sense within the real-world context the problem describes. For example, an age, a length, or a positive quantity in the past cannot be negative — if one root produces a negative age or a negative length, it must be rejected as inadmissible, even though it is mathematically correct as a solution to the equation itself. Checking both roots against the real-world context is a required, frequently-tested step, not an optional one.

How do I know whether to use factorization or the quadratic formula to solve an equation?

Try factorization first by looking for two numbers that multiply to give a×c and add to give b — if these numbers are easy to spot, factorization is faster. If the equation does not factor neatly with integer or simple fractional numbers, or if you cannot quickly identify the factor pair, switch to the quadratic formula, which works for every quadratic equation regardless of whether it factors nicely.

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