Question 1

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

Accepted Answer

**Answer: (a) $3,4$.** $x^2-7x+12=(x-3)(x-4)=0$, so $x=3$ or $x=4.$

Question 2

The discriminant of $2x^2-3x+1=0$ is:

Accepted Answer

**Answer: (a) $1$.** $D=b^2-4ac=(-3)^2-4(2)(1)=9-8=1.$

Question 3

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

Accepted Answer

**Answer: (a) Real and equal.** $D=4^2-4(1)(4)=16-16=0$, so the roots are real and equal ($x=-2,-2$).

Question 4

The equation $4x^2+kx+9=0$ has equal roots when $k$ equals:

Accepted Answer

**Answer: (a) $\pm12$.** Equal roots need $D=0$: $k^2-4(4)(9)=0\Rightarrow k^2=144\Rightarrow k=\pm12.$

Question 5

**Assertion (A):** The equation $x^2+x+1=0$ has no real roots. **Reason (R):** A quadratic equation has no real roots when its discriminant $b^2-4ac<0$.

Accepted Answer

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

Question 6

Solve by factorisation: $6x^2-x-2=0$.

Accepted Answer

Split the middle term ($6	imes-2=-12$; $-4$ and $+3$): $6x^2-4x+3x-2=0$ $2x(3x-2)+1(3x-2)=0$ $(3x-2)(2x+1)=0.$ So $x=\dfrac{2}{3}$ or $x=-\dfrac{1}{2}.$

Question 7

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

Accepted Answer

$D=b^2-4ac=(-2)^2-4(3)(1)=4-12=-8.$ Since $D=-8<0$, the equation has no real roots.

Question 8

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

Accepted Answer

Here $a=1$, $b=-4$, $c=-2$. $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{4\pm\sqrt{16+8}}{2}=\dfrac{4\pm\sqrt{24}}{2}.$ $x=\dfrac{4+4.899}{2}=\dfrac{8.899}{2}=4.45$ or $x=\dfrac{4-4.899}{2}=\dfrac{-0.899}{2}=-0.45.$ So $x\approx4.45$ or $x\approx-0.45.$

Question 9

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

Accepted Answer

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

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

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

Question 10

The sum of a number and its reciprocal is $2\dfrac{1}{6}$. Find the number.

Accepted Answer

Let the number be $x$. Then $x+\dfrac{1}{x}=2\dfrac16=\dfrac{13}{6}.$ Multiply by $6x$: $6x^2+6=13x\Rightarrow6x^2-13x+6=0.$ Split ($6	imes6=36$; $-9$ and $-4$): $6x^2-9x-4x+6=0\Rightarrow3x(2x-3)-2(2x-3)=0\Rightarrow(2x-3)(3x-2)=0.$ So $x=\dfrac32$ or $x=\dfrac23.$

Question 11

An express train makes a run of $240$ km at a certain speed. Another train, whose speed is $12$ km/h less, takes $1$ hour longer to cover the same distance. Find the speed of the express train.

Accepted Answer

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

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

Given $\dfrac{240}{x-12}-\dfrac{240}{x}=1.$

$240x-240(x-12)=x(x-12)$

$240\times12=x^2-12x\Rightarrow x^2-12x-2880=0.$

$x=\dfrac{12\pm\sqrt{144+11520}}{2}=\dfrac{12\pm\sqrt{11664}}{2}=\dfrac{12\pm108}{2}.$

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

Question 12

The hypotenuse of a right-angled triangle is $13$ cm. If one of the remaining two sides is $7$ cm longer than the other, find the lengths of these two sides.

Accepted Answer

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

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

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

$2x^2+14x-120=0\Rightarrow x^2+7x-60=0.$

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

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

Question 13

A rectangular garden has its length $5$ m more than its breadth, and its area is $84\ \text{m}^2$ .

(i) Taking the breadth as $x$ 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.

Accepted Answer

**(i)** Length $=(x+5)$ m, so area $=x(x+5)=84\Rightarrow x^2+5x-84=0.$ **(ii)** $(x+12)(x-7)=0\Rightarrow x=7$ or $x=-12$. Length cannot be negative, so $x=7.$ **(iii)** Breadth $=7$ m and length $=7+5=12$ m. **(iv)** Perimeter $=2(	ext{length}+	ext{breadth})=2(12+7)=2	imes19=38$ m.

Question 14

Are these Quadratic Equations important questions free?

Accepted Answer

Yes. All 13 ICSE Class 10 Maths important questions for Quadratic Equations are free, with full model answers and no login required.

Question 15

Do these Quadratic Equations questions follow the latest ICSE syllabus?

Accepted Answer

Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Quadratic Equations important questions?

Accepted Answer

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.

Question 17

What types of questions are covered for Quadratic Equations?

Accepted Answer

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 ICSE paper is covered.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

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