SEBA Class 10 Maths Exercise 4.2 Solutions: Quadratic Equations | New Book 2026
Get free solutions to SEBA Class 10 Maths Exercise 4.2 Solutions. This Exercise 4.2 is from SEBA’s new book, 2026. We have solved all the questions in a simple way so that you can understand the concepts and get full marks in your upcoming HSLC examination.
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Exercise 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) $x^{2}-3x-10=0$
$\mathbf{Sol^n.}$
Given:$x^{2}-3x-10=0$
$\Rightarrow x^{2}-(5-2)x-10=0$
$\Rightarrow x^{2}-5x+2x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow (x-5)(x+2)=0$
Either, $x-5=0$ or, $x+2=0$
$\Rightarrow x=5$ or $ x=-2$
Therefore, the roots of the given equation are $5, -2$.
(ii) $2x^{2}+x-6=0$
$\mathbf{Sol^n.}$
We have: $2x^{2}+x-6=0$
$\Rightarrow 2x^{2}+(4-3)x-6=0$
$\Rightarrow 2x^{2}+4x-3x-6=0$
$\Rightarrow 2x(x+2)-3(x+2)=0$
$\Rightarrow (x+2)(2x-3)=0$
Either, $x+2=0$
$\Rightarrow x=-2$
or, $2x-3=0$
$\Rightarrow 2x=3$
$\Rightarrow x=\frac{3}{2}$
Therefore, the roots of the given equation are $-2, \frac{3}{2}$.
(iii) $\sqrt{2}x^{2}+7x+5\sqrt{2}=0$
$\mathbf{Sol^n.}$
Given:$\sqrt{2}x^{2}+7x+5\sqrt{2}=0$
$\Rightarrow \sqrt{2}x^{2}+(5+2)x+5\sqrt{2}=0$
$\Rightarrow \sqrt{2}x^{2}+5x+2x+5\sqrt{2}=0$
$\Rightarrow x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$
$\Rightarrow (\sqrt{2}x+5)(x+\sqrt{2})=0$
Either, $\sqrt{2}x+5=0$
$\Rightarrow \sqrt{2}x=-5$
$\Rightarrow x=-\frac{5}{\sqrt{2}}$
or, $x+\sqrt{2}=0$
$\Rightarrow x=-\sqrt{2}$
Therefore, the roots of the given equation are $-\frac{5}{\sqrt{2}}, -\sqrt{2}$.
(iv) $2x^{2}-x+\frac{1}{8}=0$
$\mathbf{Sol^n.}$
Given: $2x^{2}-x+\frac{1}{8}=0$
$\Rightarrow \frac{16x^{2}-8x+1}{8}=0$
$\Rightarrow 16x^{2}-8x+1=0$
$\Rightarrow 16x^{2}-(4+4)x+1=0$
$\Rightarrow 16x^{2}-4x-4x+1=0$
$\Rightarrow 4x(4x-1)-1(4x-1)=0$
$\Rightarrow (4x-1)(4x-1)=0$
Either, $4x-1=0$ or $4x-1=0$
$\Rightarrow x=\frac{1}{4}$ or $ x=\frac{1}{4}$
Therefore, the roots of the given equation are $\frac{1}{4}, \frac{1}{4}$.
(v) $100x^{2}-20x+1=0$
$\mathbf{Sol^n.}$
Given: $100x^{2}-20x+1=0$
$\Rightarrow 100x^{2}-(10+10)x+1=0$
$\Rightarrow 100x^{2}-10x-10x+1=0$
$\Rightarrow 10x(10x-1)-1(10x-1)=0$
$\Rightarrow (10x-1)(10x-1)=0$
Either, $10x-1=0$ or $10x-1=0$
$\Rightarrow x=\frac{1}{10}$ or $ x=\frac{1}{10}$
Therefore, the roots of the given equation are $\frac{1}{10}, \frac{1}{10}$.
(vi) $2x^{2}-7x+6=0$
$\mathbf{Sol^n.}$
Given: $2x^{2}-7x+6=0$
$\Rightarrow 2x^{2}-(4+3)x+6=0$
$\Rightarrow 2x^{2}-4x-3x+6=0$
$\Rightarrow 2x(x-2)-3(x-2)=0$
$\Rightarrow (x-2)(2x-3)=0$
Either, $x-2=0$ or, $2x-3=0$
$\Rightarrow x=2$ or $ x=\frac{3}{2}$
Therefore, the roots of the given equation are $2, \frac{3}{2}$.
(vii) $x^{2}-10x-96=0$
$\mathbf{Sol^n.}$
Given: $x^{2}-10x-96=0$
$\Rightarrow x^{2}-(16-6)x-96=0$
$\Rightarrow x^{2}-16x+6x-96=0$
$\Rightarrow x(x-16)+6(x-16)=0$
$\Rightarrow (x-16)(x+6)=0$
Either, $x-16=0$ or, $x+6=0$
$\Rightarrow x=16$ or $ x=-6$
Therefore, the roots of the given equation are $16, -6$.
(viii) $\sqrt{3}x^{2}+10x+7\sqrt{3}=0$
$\mathbf{Sol^n.}$
Given: $\sqrt{3}x^{2}+10x+7\sqrt{3}=0$
$\Rightarrow \sqrt{3}x^{2}+(7+3)x+7\sqrt{3}=0$
$\Rightarrow \sqrt{3}x^{2}+7x+3x+7\sqrt{3}=0$
$\Rightarrow x(\sqrt{3}x+7)+\sqrt{3}(\sqrt{3}x+7)=0$
$\Rightarrow (\sqrt{3}x+7)(x+\sqrt{3})=0$
Either, $\sqrt{3}x+7=0$ or, $x+\sqrt{3}=0$
$\Rightarrow x=\frac{-7}{\sqrt{3}}$ or $ x=-\sqrt{3}$
Therefore, the roots of the given equation are $\frac{-7}{\sqrt{3}}, -\sqrt{3}$.
(ix) $x^{2}+2\sqrt{2}x+2=0$
$\mathbf{Sol^n.}$
Given: $x^{2}+2\sqrt{2}x+2=0$
$\Rightarrow (x)^{2}+2\times \sqrt{2}\times x+(\sqrt{2})^{2}=0$
$\Rightarrow (x+\sqrt{2})^{2}=0$
$\Rightarrow (x+\sqrt{2})(x+\sqrt{2})=0$
Either, $x+\sqrt{2}=0$ or, $x+\sqrt{2}=0$
$\Rightarrow x=-\sqrt{2}$ or $ x=-\sqrt{2}$
Therefore, the roots of the given equation are $-\sqrt{2}, -\sqrt{2}$.
(x) $14x+5-3x^{2}=0$
$\mathbf{Sol^n.}$
Given: $14x+5-3x^{2}=0$
$\Rightarrow -3x^{2}+14x+5=0$
$\Rightarrow 3x^{2}-14x-5=0$
$\Rightarrow 3x^{2}-(15-1)x-5=0$
$\Rightarrow 3x^{2}-15x+x-5=0$
$\Rightarrow 3x(x-5)+1(x-5)=0$
$\Rightarrow (x-5)(3x+1)=0$
Either, $x-5=0$ or, $3x+1=0$
$\Rightarrow x=5$ or $ x=\frac{-1}{3}$
Therefore, the roots of the given equation are $5, \frac{-1}{3}$.
2. Solve the problems given in Example 1.
(i) John and Jayanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
$\mathbf{Sol^n.}$
Let the number of marbles John had be $x$.
Then, the number of marbles Jayanti had $= (45-x)$.
It is given that both of them had lost 5 marbles each.
$\therefore$ The number of marbles left with John $= (x-5)$ and,
The number of marbles left with Jayanti $= (45-x-5)$ i.e. $(40-x)$.
According to question,
$(x-5)(40-x)=124$
$\Rightarrow (x-5)(40-x)-124=0$
$\Rightarrow x(40-x)-5(40-x)-124=0$
$\Rightarrow 40x-x^{2}-200+5x-124=0$
$\Rightarrow -x^{2}+45x-324=0$
$\Rightarrow -(x^{2}-45x+324)=0$
$\Rightarrow x^{2}-45x+324=0$
$\Rightarrow x^{2}-(36+9)x+324=0$
$\Rightarrow x^{2}-36x-9x+324=0$
$\Rightarrow x(x-36)-9(x-36)=0$ $\Rightarrow (x-36)(x-9)=0$
Either, $x-36=0$ or, $x-9=0$
$\Rightarrow x=36$ or $x=9$
Therefore, the number of marbles they had to start with $9$ and $36$.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.
$\mathbf{Sol^n.}$
Let the number of toys produced on that day be $x$.
Then, the cost of production on that day $=\mathbb{R}(55-x)$.
According to question,
$x(55-x)=750$
$\Rightarrow 55x-x^{2}-750=0$
$\Rightarrow -x^{2}+55x-750=0$
$\Rightarrow -(x^{2}-55x+750)=0$
$\Rightarrow x^{2}-55x+750=0$
$\Rightarrow x^{2}-(30+25)x+750=0$
$\Rightarrow x^{2}-30x-25x+750=0$
$\Rightarrow x(x-30)-25(x-30)=0$
$\Rightarrow (x-30)(x-25)=0$
Either, $x-30=0$ or, $x-25=0$
$\Rightarrow x=30$ or $ x=25$
Therefore, the number of toys produced on that particular day was either 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
$\mathbf{Sol^n.}$
Let one of the numbers be $x$.
Then another number will be $(27-x)$.
According to question,
$x(27-x)=182$
$\Rightarrow 27x-x^{2}=182$
$\Rightarrow x^{2}-27x+182=0$
$\Rightarrow x^{2}-(13+14)x+182=0$
$\Rightarrow x^{2}-13x-14x+182=0$
$\Rightarrow x(x-13)-14(x-13)=0$
$\Rightarrow (x-13)(x-14)=0$
Either, $x-13=0$ or, $x-14=0$
$\Rightarrow x=13$ or $ x=14$
Therefore, the two numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
$\mathbf{Sol^n.}$
Let the two consecutive positive integers be $x$ and $x+1$.
According to question,
$x^{2}+(x+1)^{2}=365$
$\Rightarrow x^{2}+x^{2}+2.x.1+1^{2}=365$
$\Rightarrow 2x^{2}+2x+1-365=0$
$\Rightarrow 2x^{2}+2x-364=0$
$\Rightarrow 2(x^{2}+x-182)=0$
$\Rightarrow x^{2}+x-182=0$
$\Rightarrow x^{2}+(14-13)x-182=0$
$\Rightarrow x^{2}+14x-13x-182=0$
$\Rightarrow x(x+14)-13(x+14)=0$
$\Rightarrow (x+14)(x-13)=0$
Either, $x+14=0$ or, $x-13=0$
$\Rightarrow x=-14$ or $x=13$
But, $x=-14$ is not admissible as the required integers are positive.
Therefore, two consecutive positive integers are $13$ and $13+1=14$.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
$\mathbf{Sol^n.}$
Let $ABC$ be a right-angled triangle, where $\angle B=90^{\circ}$.
Let the base be, $BC=x\text{ cm}$
Then altitude will be, $AB=(x-7)\text{ cm}$
Hypotenuse, $AC=13\text{ cm}$
By Pythagoras Theorem, we have:
$13^{2}=(x-7)^{2}+x^{2}$
$\Rightarrow 169=x^{2}-14x+49+x^{2}$
$\Rightarrow 169=2x^{2}-14x+49$
$\Rightarrow 2x^{2}-14x+49-169=0$
$\Rightarrow 2x^{2}-14x-120=0$
$\Rightarrow 2(x^{2}-7x-60)=0$
$\Rightarrow x^{2}-7x-60=0$
$\Rightarrow x^{2}-(12-5)x-60=0$
$\Rightarrow x^{2}-12x+5x-60=0$
$\Rightarrow x(x-12)+5(x-12)=0$
$\Rightarrow (x-12)(x+5)=0$
Either, $x-12=0$ or, $x+5=0$
$\Rightarrow x=12$ or $ x=-5$ [Not admissible as the length cannot be negative]
Therefore, the base of the triangle is $12\text{ cm}$ and the altitude of the triangle is $(12-7)\text{ cm}=5\text{ cm}$.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.
$\mathbf{Sol^n.}$
Let the number of articles produced on that day be $x$.
Therefore, the cost of production (in rupees) of each article that day $=2x+3$
So, the total cost of production (in rupees) on that day $=x(2x+3)$
According to question,
$x(2x+3)=90$
$\Rightarrow 2x^{2}+3x=90$
$\Rightarrow 2x^{2}+3x-90=0$
$\Rightarrow 2x^{2}+(15-12)x-90=0$
$\Rightarrow 2x^{2}+15x-12x-90=0$
$\Rightarrow x(2x+15)-6(2x+15)=0$
$\Rightarrow (2x+15)(x-6)=0$
Either, $2x+15=0$
$\Rightarrow 2x=-15$
$\Rightarrow x=-\frac{15}{2}$ [Not admissible as the number of articles cannot be negative]
or, $x-6=0$
$\Rightarrow x=6$
Therefore, the number of articles produced on that day was 6 and the cost of each article was ₹ $(2\times6+3)= \text{Rs}15$.
7. Choose the correct option:
If $x^{2}-2px+p^{2}=0$ then the value of $\frac{p}{x}$ is
(a) 0
(b) -1
(c) 1
(d) 2
$\mathbf{Sol^n.}$
Given: $x^{2}-2px+p^{2}=0$
$\Rightarrow (x-p)^{2}=0$
$\Rightarrow x-p=0$
$\Rightarrow x=p$
$\Rightarrow 1=\frac{p}{x}$
$\therefore$ Correct Option: (c) 1
8. A student has done through the following steps to find the roots of the equation $x^{2}-3x-10=0$ by factorization method:
Step 1: $x^{2}-3x-10=0$
Step 2: $x^{2}-5x-2x-10=0$
Step 3: $x(x-5)-2(x-5)=0$
Step 4: $(x-5)(x-2)=0$
Step 5: $x=5$ and $x=2$
In which step did the student make the first mistake?
(a) step:2
(b) step:3
(c) step:4
(d) step:5
$\mathbf{Sol^n.}$
Given : $x^{2}-3x-10=0$
$\Rightarrow x^{2}-(5-2)x-10=0$
$\Rightarrow x^{2}-5x+2x-10=0$
In Step 2, the student wrote $x^{2}-5x-2x-10=0$ instead of $x^{2}-5x+2x-10=0$.
$\therefore$ The student made the first mistake in step:2.
$\therefore$ Correct Option: (a) step:2
9. Find the sum and product of the roots of the quadratic equation $2x^{2}-9x+4=0$.
$\mathbf{Sol^n.}$
Given: $2x^{2}-9x+4=0$
$\Rightarrow 2x^{2}-(8+1)x+4=0$
$\Rightarrow 2x^{2}-8x-x+4=0$
$\Rightarrow 2x(x-4)-1(x-4)=0$
$\Rightarrow (x-4)(2x-1)=0$
Either, $x-4=0$ or, $2x-1=0$
$\Rightarrow x=4$ or $ x=\frac{1}{2}$
$\therefore$ The roots are $4$ and $\frac{1}{2}$.
Now,
$\text{Sum of the roots} = 4+\frac{1}{2}=\frac{8+1}{2}=\frac{9}{2}$
$\text{Product of the roots} = 4\times \frac{1}{2}=2$
10. If one root of the quadratic equation $2x^{2}+kx-6=0$ is 2 then find the value of $k$. Also find the other root.
$\mathbf{Sol^n.}$
Given: $2x^{2}+kx-6=0$
Since $x=2$ is a root of the equation, it must satisfy the equation.
Substituting $x=2$ in the given equation: $2(2)^{2}+k(2)-6=0$, we get:
$\Rightarrow 2(4)+2k-6=0$
$\Rightarrow 8+2k-6=0$
$\Rightarrow 2k+2=0$
$\Rightarrow 2k=-2$
$\Rightarrow k=-1$
Now, substituting $k=-1$ in the given quadratic equation:
$2x^{2}-x-6=0$
$\Rightarrow 2x^{2}-(4-3)x-6=0$
$\Rightarrow 2x^{2}-4x+3x-6=0$
$\Rightarrow 2x(x-2)+3(x-2)=0$
$\Rightarrow (x-2)(2x+3)=0$
Either, $x-2=0$ or, $2x+3=0$
$\Rightarrow x=2$ or $ x=-\frac{3}{2}$
Therefore, the value of $k$ is $-1$ and the other root is $-\frac{3}{2}$.
11. A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway, shown in the figure.
The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres.
Based on the information given above, answer the following questions:
(i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway $= x\text{ m}$.
$\mathbf{Sol^n.}$ Given:
Length of the lawn $= 12\text{ m}$
Breadth of the lawn $= 10\text{ m}$
Width of the walkway on all sides $= x\text{ m}$
$\therefore$ Total length of lawn including walkway $= 12+2x$
Total breadth of lawn including walkway $= 10+2x$
Total area including walkway $= (12+2x)(10+2x)$
According to question,
$(12+2x)(10+2x)=360$
$\Rightarrow 120+24x+20x+4x^{2}=360$
$\Rightarrow 4x^{2}+44x+120-360=0$
$\Rightarrow 4x^{2}+44x-240=0$
$\Rightarrow 4(x^{2}+11x-60)=0$
$\Rightarrow x^{2}+11x-60=0$
Therefore, the required quadratic equation is $x^{2}+11x-60=0$.
(ii) Solve the quadratic equation to find the width of the walkway ‘$x$‘.
$\mathbf{Sol^n.}$
The quadratic equation is: $x^{2}+11x-60=0$
$\Rightarrow x^{2}+(15-4)x-60=0$
$\Rightarrow x^{2}+15x-4x-60=0$
$\Rightarrow x(x+15)-4(x+15)=0$
$\Rightarrow (x+15)(x-4)=0$
Either, $x+15=0$ or $x-4=0$
$\Rightarrow x=-15$ or $ x=4$
Since $x = -15$ not admissible, as width cannot be negative
Therefore, the width of the walkway ‘$x$‘ is $4\text{ m}$.
(iii) If the cost of paving the walkway at the rate of Rs. 50 per square metre is Rs. 12,000, calculate the area of the walkway.
$\mathbf{Sol^n.}$
Given:
Rate of paving $= \text{Rs. } 50 \text{ per square metre}$
Total cost of paving $= \text{Rs. } 12,000$
$\therefore$ Area of the walkway $= \frac{\text{Total Cost}}{\text{Rate}}$
$= \frac{12000}{50}$
$= 240\text{ sq. m}$
Therefore, the area of the walkway is $240\text{ square metres}$.
(iv) Find the perimeter of the lawn.
$\mathbf{Sol^n.}$
Dimensions of the lawn itself are:
Length ($l$) $= 12\text{ m}$
Breadth ($b$) $= 10\text{ m}$
$\therefore \text{Perimeter of the lawn} = 2(l+b)$
$= 2(12+10)$
$= 2(22)$
$= 44\text{ m}$
Therefore, the perimeter of the lawn is $44\text{ metres}$.