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Quadratic Equations — ICSE Class 10 Maths Important Questions

13 ICSE Class 10 Maths practice questions on Quadratic Equations, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Key concepts
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Quadratic Equations — ICSE Class 10 Maths Important Questions

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Quick answer

Standard ICSE Quadratic Equations questions are solving by factorisation and by the formula x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=-b±√b^2-4ac/2a, using the discriminant b2−4acb^2-4acb^2-4ac to judge the nature of the roots, solving equations reducible to quadratics, and forming quadratics from word problems. Nature-of-roots and 'solve correct to two decimal places' questions appear almost every year.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Quadratic Equations

Inside the ICSE Class 10 Maths chapter Quadratic Equations you solve ax2+bx+c=0ax^2+bx+c=0ax^2+bx+c=0 by factorisation and by the quadratic formula, decide the nature of the roots using the discriminant, solve equations that reduce to quadratics, express roots correct to two decimal places, and model word problems (numbers, ages, speed, geometry) as quadratic equations.

Solving by factorisationQuadratic formulaNature of roots (discriminant)Roots to given decimal placesWord problems reducible to quadratics

Key concepts & formulas

Quadratic formula

For ax2+bx+c=0 (a≠0)ax^2+bx+c=0\ (a\neq0)ax^2+bx+c=0 (a≠0), the roots are x=−b±b2−4ac2a.x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.x=-b±√b^2-4ac/2a.

Discriminant and nature of roots

D=b2−4acD=b^2-4acD=b^2-4ac. If D>0D>0D>0 the roots are real and distinct; if D=0D=0D=0 they are real and equal; if D<0D<0D<0 there are no real roots. For equal roots set D=0.D=0.D=0.

Word problems

Let the unknown be xxx, translate the conditions into a quadratic ax2+bx+c=0ax^2+bx+c=0ax^2+bx+c=0, solve, and reject any root that is impossible in the context (e.g. a negative length or age).

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The roots of x2−7x+12=0x^2-7x+12=0x^2-7x+12=0 are:

  1. (a)

    3,43,43,4

  2. (b)

    −3,−4-3,-4-3,-4

  3. (c)

    2,62,62,6

  4. (d)

    −2,−6-2,-6-2,-6

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Answer: (a) 3,43,43,4.

x2−7x+12=(x−3)(x−4)=0x^2-7x+12=(x-3)(x-4)=0x^2-7x+12=(x-3)(x-4)=0, so x=3x=3x=3 or x=4.x=4.x=4.

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Q2MCQEasy1 mark

The discriminant of 2x2−3x+1=02x^2-3x+1=02x^2-3x+1=0 is:

  1. (a)

    111

  2. (b)

    −1-1-1

  3. (c)

    171717

  4. (d)

    000

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Answer: (a) 111.

D=b2−4ac=(−3)2−4(2)(1)=9−8=1.D=b^2-4ac=(-3)^2-4(2)(1)=9-8=1.D=b^2-4ac=(-3)^2-4(2)(1)=9-8=1.

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Q3MCQModerate1 mark

The nature of the roots of x2+4x+4=0x^2+4x+4=0x^2+4x+4=0 is:

  1. (a)

    Real and equal

  2. (b)

    Real and distinct

  3. (c)

    No real roots

  4. (d)

    Imaginary and distinct

Show model answer

Answer: (a) Real and equal.

D=42−4(1)(4)=16−16=0D=4^2-4(1)(4)=16-16=0D=4^2-4(1)(4)=16-16=0, so the roots are real and equal (x=−2,−2x=-2,-2x=-2,-2).

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Q4MCQHOTS1 mark

The equation 4x2+kx+9=04x^2+kx+9=04x^2+kx+9=0 has equal roots when kkk equals:

  1. (a)

    ±12\pm12±12

  2. (b)

    121212

  3. (c)

    ±6\pm6±6

  4. (d)

    666

Show model answer

Answer: (a) ±12\pm12±12.

Equal roots need D=0D=0D=0: k2−4(4)(9)=0⇒k2=144⇒k=±12.k^2-4(4)(9)=0\Rightarrow k^2=144\Rightarrow k=\pm12.k^2-4(4)(9)=0 k^2=144 k=±12.

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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 has no real roots when its discriminant b2−4ac<0b^2-4ac<0b^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

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Answer: (a) D=12−4(1)(1)=1−4=−3<0D=1^2-4(1)(1)=1-4=-3<0D=1^2-4(1)(1)=1-4=-3<0, so there are no real roots, and R is the correct reason.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Solve by factorisation: 6x2−x−2=06x^2-x-2=06x^2-x-2=0.

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Split the middle term (6×−2=−126\times-2=-126×-2=-12; −4-4-4 and +3+3+3):

6x2−4x+3x−2=06x^2-4x+3x-2=06x^2-4x+3x-2=0

2x(3x−2)+1(3x−2)=02x(3x-2)+1(3x-2)=02x(3x-2)+1(3x-2)=0

(3x−2)(2x+1)=0.(3x-2)(2x+1)=0.(3x-2)(2x+1)=0.

So x=23x=\dfrac{2}{3}x=2/3 or x=−12.x=-\dfrac{1}{2}.x=-1/2.

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Q7Very ShortModerate2 marks

Find the discriminant of 3x2−2x+1=03x^2-2x+1=03x^2-2x+1=0 and hence state the nature of its roots.

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D=b2−4ac=(−2)2−4(3)(1)=4−12=−8.D=b^2-4ac=(-2)^2-4(3)(1)=4-12=-8.D=b^2-4ac=(-2)^2-4(3)(1)=4-12=-8.

Since D=−8<0D=-8<0D=-8<0, the equation has no real roots.

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Short answer questions (3 marks)

Q8Short AnswerEasy3 marks

Solve x2−4x−2=0x^2-4x-2=0x^2-4x-2=0 using the quadratic formula, giving the roots correct to two decimal places. (Take 24=4.899\sqrt{24}=4.899√24=4.899.)

Show model answer

Here a=1a=1a=1, b=−4b=-4b=-4, c=−2c=-2c=-2.

x=−b±b2−4ac2a=4±16+82=4±242.x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{4\pm\sqrt{16+8}}{2}=\dfrac{4\pm\sqrt{24}}{2}.x=-b±√b^2-4ac/2a=4±√16+8/2=4±√24/2.

x=4+4.8992=8.8992=4.45x=\dfrac{4+4.899}{2}=\dfrac{8.899}{2}=4.45x=4+4.899/2=8.899/2=4.45 or x=4−4.8992=−0.8992=−0.45.x=\dfrac{4-4.899}{2}=\dfrac{-0.899}{2}=-0.45.x=4-4.899/2=-0.899/2=-0.45.

So x≈4.45x\approx4.45x4.45 or x≈−0.45.x\approx-0.45.x-0.45.

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Q9Short AnswerModerate3 marks

If x=2x=2x=2 is a root of the equation kx2+2x−6=0kx^2+2x-6=0kx^2+2x-6=0, find the value of kkk and hence the other root.

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Since x=2x=2x=2 is a root, substitute: k(2)2+2(2)−6=0⇒4k+4−6=0⇒4k=2⇒k=12.k(2)^2+2(2)-6=0\Rightarrow4k+4-6=0\Rightarrow4k=2\Rightarrow k=\dfrac12.k(2)^2+2(2)-6=04k+4-6=04k=2 k=12.

The equation becomes 12x2+2x−6=0\dfrac12x^2+2x-6=012x^2+2x-6=0, i.e. x2+4x−12=0.x^2+4x-12=0.x^2+4x-12=0.

(x+6)(x−2)=0⇒x=−6(x+6)(x-2)=0\Rightarrow x=-6(x+6)(x-2)=0 x=-6 or x=2x=2x=2. The other root is x=−6.x=-6.x=-6.

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Q10Short AnswerHOTS3 marks

The sum of a number and its reciprocal is 2162\dfrac{1}{6}21/6. Find the number.

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Let the number be xxx. Then x+1x=216=136.x+\dfrac{1}{x}=2\dfrac16=\dfrac{13}{6}.x+1/x=216=13/6.

Multiply by 6x6x6x: 6x2+6=13x⇒6x2−13x+6=0.6x^2+6=13x\Rightarrow6x^2-13x+6=0.6x^2+6=13x6x^2-13x+6=0.

Split (6×6=366\times6=366×6=36; −9-9-9 and −4-4-4): 6x2−9x−4x+6=0⇒3x(2x−3)−2(2x−3)=0⇒(2x−3)(3x−2)=0.6x^2-9x-4x+6=0\Rightarrow3x(2x-3)-2(2x-3)=0\Rightarrow(2x-3)(3x-2)=0.6x^2-9x-4x+6=03x(2x-3)-2(2x-3)=0(2x-3)(3x-2)=0.

So x=32x=\dfrac32x=32 or x=23.x=\dfrac23.x=23.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

An express train makes a run of 240240240 km at a certain speed. Another train, whose speed is 121212 km/h less, takes 111 hour longer to cover the same distance. Find the speed of the express train.

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Let the speed of the express train be xxx km/h. Time taken =240x=\dfrac{240}{x}=240/x hours.

The slower train has speed (x−12)(x-12)(x-12) km/h and time 240x−12\dfrac{240}{x-12}240/x-12 hours.

Given 240x−12−240x=1.\dfrac{240}{x-12}-\dfrac{240}{x}=1.240/x-12-240/x=1.

240x−240(x−12)=x(x−12)240x-240(x-12)=x(x-12)240x-240(x-12)=x(x-12)

240×12=x2−12x⇒x2−12x−2880=0.240\times12=x^2-12x\Rightarrow x^2-12x-2880=0.240×12=x^2-12x x^2-12x-2880=0.

x=12±144+115202=12±116642=12±1082.x=\dfrac{12\pm\sqrt{144+11520}}{2}=\dfrac{12\pm\sqrt{11664}}{2}=\dfrac{12\pm108}{2}.x=12±√144+11520/2=12±√11664/2=12±108/2.

x=60x=60x=60 or x=−48x=-48x=-48. Rejecting the negative speed, the express train's speed is 606060 km/h.

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Q12Long AnswerHOTS5 marks

The hypotenuse of a right-angled triangle is 131313 cm. If one of the remaining two sides is 777 cm longer than the other, find the lengths of these two sides.

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Let the shorter side be xxx cm; then the other side is (x+7)(x+7)(x+7) cm.

By Pythagoras' theorem: x2+(x+7)2=132.x^2+(x+7)^2=13^2.x^2+(x+7)^2=13^2.

x2+x2+14x+49=169x^2+x^2+14x+49=169x^2+x^2+14x+49=169

2x2+14x−120=0⇒x2+7x−60=0.2x^2+14x-120=0\Rightarrow x^2+7x-60=0.2x^2+14x-120=0 x^2+7x-60=0.

(x+12)(x−5)=0⇒x=5(x+12)(x-5)=0\Rightarrow x=5(x+12)(x-5)=0 x=5 or x=−12.x=-12.x=-12.

Rejecting the negative length, x=5x=5x=5. The two sides are 555 cm and 5+7=125+7=125+7=12 cm.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A rectangular garden has its length 555 m more than its breadth, and its area is 84 m284\ \text{m}^284 m^2.

(i) Taking the breadth as xxx m, form a quadratic equation.

(ii) Solve it to find the breadth.

(iii) State the length and breadth of the garden.

(iv) Find the perimeter of the garden.

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(i) Length =(x+5)=(x+5)=(x+5) m, so area =x(x+5)=84⇒x2+5x−84=0.=x(x+5)=84\Rightarrow x^2+5x-84=0.=x(x+5)=84 x^2+5x-84=0.

(ii) (x+12)(x−7)=0⇒x=7(x+12)(x-7)=0\Rightarrow x=7(x+12)(x-7)=0 x=7 or x=−12x=-12x=-12. Length cannot be negative, so x=7.x=7.x=7.

(iii) Breadth =7=7=7 m and length =7+5=12=7+5=12=7+5=12 m.

(iv) Perimeter =2(length+breadth)=2(12+7)=2×19=38=2(\text{length}+\text{breadth})=2(12+7)=2\times19=38=2(length+breadth)=2(12+7)=2×19=38 m.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Quadratic Equations questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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