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Quadratic Equations — CBSE Class 10 Maths Important Questions

13 hand-picked CBSE Class 10 Maths important questions for Quadratic Equations, each with a full model answer — the formats and topics most likely to appear in your board exam.

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Quadratic Equations — CBSE Class 10 Maths Important Questions

Every Root, Every Method

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Quick answer

Board favourites are the discriminant D=b2−4acD=b^2-4acD=b^2-4ac and the nature of roots, finding an unknown constant for equal or real roots, solving by factorisation and the quadratic formula, and 555-mark word problems on speed–time, areas, numbers and work done.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Quadratic Equations

A quadratic ax2+bx+c=0 (a≠0)ax^2+bx+c=0\ (a\neq0)ax^2+bx+c=0 (a≠0) has roots x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=-b±√b^2-4ac/2a. The sign of the discriminant D=b2−4acD=b^2-4acD=b^2-4ac decides whether the roots are real and distinct, real and equal, or non-real — the single most tested idea in this chapter.

Standard form and checking a rootSolving by factorisationThe quadratic formulaDiscriminant and nature of rootsEqual-roots conditionsWord problems

Key concepts & formulas

Nature of roots

For ax2+bx+c=0ax^2+bx+c=0ax^2+bx+c=0,  D=b2−4ac\ D=b^2-4acD=b^2-4ac. If D>0D>0D>0: two distinct real roots; if D=0D=0D=0: two equal real roots (x=−b2a)\left(x=-\dfrac{b}{2a}\right)(x=-b/2a); if D<0D<0D<0: no real roots.

The quadratic formula

x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=-b±√b^2-4ac/2a, valid whenever D≥0D\ge0D≥0. Always simplify the surd fully before writing the final roots.

Equal roots

A quadratic has equal roots exactly when D=0D=0D=0. This yields an equation in the unknown constant — the commonest 222–333 mark question in the chapter.

Word-problem strategy

Let the unknown be xxx, translate the condition into a quadratic, solve it, and reject any root that is not physically valid (a negative length, speed, age or time).

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The discriminant of the quadratic equation 2x2−4x+3=02x^2-4x+3=02x^2-4x+3=0 is:

  1. (a)

    −8-8-8

  2. (b)

    888

  3. (c)

    404040

  4. (d)

    −40-40-40

Show model answer

Answer: (a) −8-8-8 — D=b2−4ac=(−4)2−4(2)(3)=16−24=−8D=b^2-4ac=(-4)^2-4(2)(3)=16-24=-8D=b^2-4ac=(-4)^2-4(2)(3)=16-24=-8. Since D<0D<0D<0, the equation has no real roots.

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Q2MCQEasy1 mark

The roots of the quadratic equation x2−3x−10=0x^2-3x-10=0x^2-3x-10=0 are:

  1. (a)

    5, −25,\ -25, -2

  2. (b)

    −5, 2-5,\ 2-5, 2

  3. (c)

    5, 25,\ 25, 2

  4. (d)

    −5, −2-5,\ -2-5, -2

Show model answer

Answer: (a) 5, −25,\ -25, -2 — x2−3x−10=(x−5)(x+2)=0x^2-3x-10=(x-5)(x+2)=0x^2-3x-10=(x-5)(x+2)=0, so x=5x=5x=5 or x=−2x=-2x=-2.

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Q3MCQModerate1 mark

The value(s) of kkk for which 2x2+kx+3=02x^2+kx+3=02x^2+kx+3=0 has two equal roots are:

  1. (a)

    ±26\pm2\sqrt6±26

  2. (b)

    ±6\pm\sqrt6±6

  3. (c)

    ±6\pm6±6

  4. (d)

    ±23\pm2\sqrt3±23

Show model answer

Answer: (a) ±26\pm2\sqrt6±26 — For equal roots D=0D=0D=0:  k2−4(2)(3)=0⇒k2=24⇒k=±26\ k^2-4(2)(3)=0\Rightarrow k^2=24\Rightarrow k=\pm2\sqrt6k^2-4(2)(3)=0 k^2=24 k=±26.

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Q4MCQModerate1 mark

If −5-5-5 is a root of the quadratic equation 2x2+px−15=02x^2+px-15=02x^2+px-15=0, then the value of ppp is:

  1. (a)

    777

  2. (b)

    −7-7-7

  3. (c)

    333

  4. (d)

    555

Show model answer

Answer: (a) 777 — Substituting x=−5x=-5x=-5:  2(25)+p(−5)−15=0⇒50−5p−15=0⇒5p=35⇒p=7\ 2(25)+p(-5)-15=0\Rightarrow 50-5p-15=0\Rightarrow 5p=35\Rightarrow p=72(25)+p(-5)-15=0 50-5p-15=0 5p=35 p=7.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The equation x2+x+1=0x^2+x+1=0x^2+x+1=0 has no real roots.

Reason (R): A quadratic equation ax2+bx+c=0ax^2+bx+c=0ax^2+bx+c=0 has real roots only if b2−4ac≥0b^2-4ac\ge0b^2-4ac≥0.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) Both are true and R explains A. For x2+x+1=0x^2+x+1=0x^2+x+1=0,  D=12−4(1)(1)=−3<0\ D=1^2-4(1)(1)=-3<0D=1^2-4(1)(1)=-3<0, so it has no real roots — exactly the condition in R.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the discriminant of 2x2−3x+5=02x^2-3x+5=02x^2-3x+5=0 and hence comment on the nature of its roots.

Show model answer

D=b2−4ac=(−3)2−4(2)(5)=9−40=−31D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31. Since D<0D<0D<0, the equation has no real roots (the roots are non-real).

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Q7Very ShortEasy2 marks

Find the value(s) of kkk for which the quadratic equation 2x2+kx+2=02x^2+kx+2=02x^2+kx+2=0 has equal roots.

Show model answer

Equal roots ⇒D=0\Rightarrow D=0D=0:  k2−4(2)(2)=0⇒k2=16⇒k=±4\ k^2-4(2)(2)=0\Rightarrow k^2=16\Rightarrow k=\pm4k^2-4(2)(2)=0 k^2=16 k=±4.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Find two consecutive positive integers, the sum of whose squares is 365365365.

Show model answer

Let the integers be xxx and x+1x+1x+1.

x2+(x+1)2=365⇒2x2+2x+1=365⇒2x2+2x−364=0⇒x2+x−182=0.x^2+(x+1)^2=365\Rightarrow 2x^2+2x+1=365\Rightarrow 2x^2+2x-364=0\Rightarrow x^2+x-182=0.x^2+(x+1)^2=365 2x^2+2x+1=365 2x^2+2x-364=0 x^2+x-182=0.

(x+14)(x−13)=0⇒x=13(x+14)(x-13)=0\Rightarrow x=13(x+14)(x-13)=0 x=13 (rejecting x=−14x=-14x=-14 since the integers are positive).

The integers are 131313 and 141414.

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Q9Short AnswerHOTS3 marks

The difference of the squares of two numbers is 180180180. The square of the smaller number is 888 times the larger number. Find the two numbers.

Show model answer

Let the larger number be xxx and the smaller be yyy. Given x2−y2=180x^2-y^2=180x^2-y^2=180 and y2=8xy^2=8xy^2=8x.

Substituting: x2−8x=180⇒x2−8x−180=0⇒(x−18)(x+10)=0⇒x=18x^2-8x=180\Rightarrow x^2-8x-180=0\Rightarrow (x-18)(x+10)=0\Rightarrow x=18x^2-8x=180 x^2-8x-180=0 (x-18)(x+10)=0 x=18 (taking x>0x>0x>0 so that y2=8x>0y^2=8x>0y^2=8x>0).

Then y2=8(18)=144⇒y=±12y^2=8(18)=144\Rightarrow y=\pm12y^2=8(18)=144 y=±12.

The numbers are 181818 and 121212 (or 181818 and −12-12-12).

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Q10Short AnswerHOTS3 marks

If the roots of the equation (b−c)x2+(c−a)x+(a−b)=0(b-c)x^2+(c-a)x+(a-b)=0(b-c)x^2+(c-a)x+(a-b)=0 are equal, prove that 2b=a+c2b=a+c2b=a+c.

Show model answer

The coefficients sum to zero: (b−c)+(c−a)+(a−b)=0(b-c)+(c-a)+(a-b)=0(b-c)+(c-a)+(a-b)=0, so x=1x=1x=1 is always a root. For equal roots both roots must equal 111, hence the product of the roots is 111:
a−bb−c=1⇒a−b=b−c⇒a+c=2b.\frac{a-b}{b-c}=1\Rightarrow a-b=b-c\Rightarrow a+c=2b.a-b/b-c=1 a-b=b-c a+c=2b.
Hence proved.

(Equivalently, D=(c−a)2−4(b−c)(a−b)=0D=(c-a)^2-4(b-c)(a-b)=0D=(c-a)^2-4(b-c)(a-b)=0 simplifies to (a+c−2b)2=0(a+c-2b)^2=0(a+c-2b)^2=0.)

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

A train travels 360360360 km at a uniform speed. If the speed had been 555 km/h more, it would have taken 111 hour less for the same journey. Find the speed of the train.

Show model answer

Let the speed be xxx km/h, so the time taken is 360x\dfrac{360}{x}360/x hours.

360x−360x+5=1⇒360((x+5)−xx(x+5))=1⇒1800x2+5x=1.\dfrac{360}{x}-\dfrac{360}{x+5}=1\Rightarrow 360\left(\dfrac{(x+5)-x}{x(x+5)}\right)=1\Rightarrow \dfrac{1800}{x^2+5x}=1.360/x-360/x+5=1 360((x+5)-x/x(x+5))=1 1800/x^2+5x=1.

x2+5x−1800=0.x^2+5x-1800=0.x^2+5x-1800=0. Here D=25+7200=7225=852D=25+7200=7225=85^2D=25+7200=7225=85^2, so x=−5±852x=\dfrac{-5\pm85}{2}x=-5±85/2.

Taking the positive value, x=802=40x=\dfrac{80}{2}=40x=80/2=40.

The speed of the train is 404040 km/h.

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Q12Long AnswerHOTS5 marks

Two water taps together can fill a tank in 9389\dfrac{3}{8}93/8 hours. The tap of larger diameter takes 101010 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Show model answer

Let the smaller tap take xxx hours; then the larger takes (x−10)(x-10)(x-10) hours. Together they fill the tank in 758\dfrac{75}{8}75/8 hours, so
1x+1x−10=875.\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}.1/x+1/x-10=8/75.
(x−10)+xx(x−10)=875⇒75(2x−10)=8(x2−10x)⇒8x2−230x+750=0⇒4x2−115x+375=0.\dfrac{(x-10)+x}{x(x-10)}=\dfrac{8}{75}\Rightarrow 75(2x-10)=8(x^2-10x)\Rightarrow 8x^2-230x+750=0\Rightarrow 4x^2-115x+375=0.(x-10)+x/x(x-10)=8/75 75(2x-10)=8(x^2-10x) 8x^2-230x+750=0 4x^2-115x+375=0.

D=1152−4(4)(375)=13225−6000=7225=852D=115^2-4(4)(375)=13225-6000=7225=85^2D=115^2-4(4)(375)=13225-6000=7225=85^2, so x=115±858x=\dfrac{115\pm85}{8}x=115±85/8, giving x=25x=25x=25 or x=3.75x=3.75x=3.75.

Since x−10x-10x-10 must be positive, reject x=3.75x=3.75x=3.75. Thus x=25x=25x=25.

Smaller tap: 252525 hours; larger tap: 151515 hours.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A landscaper is designing a rectangular flower bed. Its area is to be 528 m2528\ \text{m}^2528 m^2 and its length is to be one metre more than twice its breadth.

(i) Taking the breadth as xxx m, form a quadratic equation for the situation.
(ii) Find the value of the discriminant.
(iii) Find the length and breadth of the flower bed.

Show model answer

(i) Length =(2x+1)=(2x+1)=(2x+1) m, so x(2x+1)=528⇒2x2+x−528=0x(2x+1)=528\Rightarrow 2x^2+x-528=0x(2x+1)=528 2x^2+x-528=0.

(ii) D=b2−4ac=12−4(2)(−528)=1+4224=4225=652D=b^2-4ac=1^2-4(2)(-528)=1+4224=4225=65^2D=b^2-4ac=1^2-4(2)(-528)=1+4224=4225=65^2.

(iii) x=−1±654x=\dfrac{-1\pm65}{4}x=-1±65/4. Taking the positive root, x=644=16x=\dfrac{64}{4}=16x=64/4=16. So breadth =16=16=16 m and length =2(16)+1=33=2(16)+1=33=2(16)+1=33 m. (Check: 16×33=52816\times33=52816×33=528 ✓.)

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

Related study guide

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Factorisation, formula, completing the square — the three methods, when to use which, and the small mistakes that cost board marks.
Trigonometry Class 10 CBSE: Understand Every Formula in 1 Hour
Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • Are these Quadratic Equations important questions free?
    Yes. All 13 CBSE Class 10 Maths important questions for Quadratic Equations are free, with full model answers and no login required.
  • Do these Quadratic Equations questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.
  • How should I practise the Quadratic Equations important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Quadratic Equations?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

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