CBSE Class 10 Maths Chapter-4 Quadratic Equations – Free PDF Download
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CBSE Class 10 Maths Chapter-4 Quadratic Equations Important Questions
CBSE Class 10 Maths Important Questions Chapter 4 – Quadratic Equations
3 Mark Questions
1. Check whether the following are Quadratic Equations.
(i) (x+1)2=2(x−3)
(ii) x2−2x=(−2)(3−x)
(iii) (x−2)(x+1)=(x−1)(x+3)
(iv) (x−3)(2x+1)=x(x+5)
(v) (2x−1)(x−3)=(x+5)(x−1)
(vi) x2+3x+1=(x−2)2
(vii) (x+2)3=2x(x2−1)
(viii) x3−4x2 – x+1=(x−2)3
Ans. (i) (x+1)2=2(x−3) {(a+b)2=a2+2ab+b2}
⇒ x2+1+2x=2x– 6
⇒ x2+7=0
Here, degree of equation is 2.
Therefore, it is a Quadratic Equation.
(ii) x2−2x=(−2)(3−x)
⇒ x2−2x=−6+2x
⇒ x2−2x−2x+6=0
⇒ x2−4x+6=0
Here, degree of equation is 2.
Therefore, it is a Quadratic Equation.
(iii) (x−2)(x+1)=(x−1)(x+3)
⇒ x2+x−2x– 2=x2+3x– x– 3=0
⇒ x2+x−2x– 2−x2−3x+x+3=0
⇒ x−2x– 2−3x+x+3=0
⇒ −3x+1=0
Here, degree of equation is 1.
Therefore, it is not a Quadratic Equation.
(iv) (x−3)(2x+1)=x(x+5)
⇒ 2x2+x−6x– 3=x2+5x
⇒ 2x2+x−6x– 3−x2−5x=0
⇒ x2−10x– 3=0
Here, degree of equation is 2.
Therefore, it is a quadratic equation.
(v) (2x−1)(x−3)=(x+5)(x−1)
⇒ 2x2−6x– x+3=x2 – x+5x– 5
⇒ 2x2−7x+3−x2+x−5x+5=0
⇒ x2−11x+8=0
Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.
(vi) x2+3x+1=(x−2)2 {(a−b)2=a2−2ab+b2}
⇒ x2+3x+1=x2+4−4x
⇒ x2+3x+1−x2+4x– 4=0
⇒ 7x– 3=0
Here, degree of equation is 1.
Therefore, it is not a Quadratic Equation.
(vii) (x+2)3=2x(x2−1) {(a+b)3=a3+b3+3ab(a+b)}
⇒ x3+23+3(x)(2)(x+2)=2x(x2−1)
⇒ x3+8+6x(x+2)=2x3−2x
⇒ 2x3−2x−x3 – 8−6x2−12x=0
⇒ x3−6x2−14x– 8=0
Here, degree of Equation is 3.
Therefore, it is not a quadratic Equation.
(viii) x3−4x2 – x+1=(x−2)3 {(a−b)3=a3−b3−3ab(a−b)}
⇒ x3−4x2 – x+1=x3−23 – 3(x)(2)(x−2)
⇒ −4x2 – x+1=−8−6x2+12x
⇒ 2x2−13x+9=0
Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.
2. Represent the following situations in the form of Quadratic Equations:
(i) The area of rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive numbers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at uniform speed. If, the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find speed of the train.
Ans. (i) We are given that area of a rectangular plot is 528 m2.
Let breadth of rectangular plot be x metres
Length is one more than twice its breadth.
Therefore, length of rectangular plot is (2x+1) metres
Area of rectangle = length × breadth
⇒ 528=x(2x+1)
⇒ 528=2x2+x
⇒ 2x2+x– 528=0
This is a Quadratic Equation.
(ii) Let two consecutive numbers be x and (x+1).
It is given that x(x+1)=306
⇒ x2+x=306
⇒ x2+x– 306=0
This is a Quadratic Equation.
(iii) Let present age of Rohan = x years
Let present age of Rohan’s mother = (x +26) years
Age of Rohan after 3 years = (x+3) years
Age of Rohan’s mother after 3 years = x+26+3 = (x+29) years
According to given condition:
(x+3)(x+29)=360
⇒ x2+29x+3x+87=360
⇒ x2+32x– 273=0
This is a Quadratic Equation.
(iv) Let speed of train be x km/h
Time taken by train to cover 480 km = 480x hours
If, speed had been 8km/h less then time taken would be (480x−8) hours
According to given condition, if speed had been 8km/h less then time taken is 3 hours less.
Therefore, 480x– 8=480x+3
⇒ 480(1x– 8−1x)=3
⇒ 480(x– x+8) (x) (x−8)=3
⇒ 480×8=3(x)(x−8)
⇒ 3840=3x2−24x
⇒ 3x2−24x– 3840=0
Dividing equation by 3, we get
⇒ x2−8x– 1280=0
This is a Quadratic Equation.
3. Find the roots of the following Quadratic Equations by factorization.
(i) x2−3x– 10=0
(ii) 2x2+x– 6=0
(iii) 
(iv) 
(v) 100x2−20x+1=0
Ans. (i) x2−3x– 10=0
⇒ x2−5x+2x– 10=0
⇒ x(x−5)+2(x−5)=0
⇒ (x−5)(x+2)=0
⇒ x=5,−2
(ii) 2x2+x– 6=0
⇒ 2x2+4x−3x– 6=0
⇒ 2x(x+2) – 3(x+2)=0
⇒ (2x−3)(x+2)=0
⇒ x=
(iii)
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
(iv) 2x2 – x+
=0
⇒ 
⇒ 16x2−8x+1=0
⇒ 16x2−4x−4x+1=0
⇒ 4x(4x−1) – 1(4x−1)=0
⇒ (4x−1)(4x−1)=0
⇒ x= ¼, ¼
(v) 100x2−20x+1=0
⇒ 100x2−10x−10x+1=0
⇒ 10x(10x−1) – 1(10x−1)=0
⇒ (10x−1)(10x−1)=0
⇒ x=
4. Find the roots of the following equations:
(i) 
(ii) 
Ans. (i) 
⇒ 
⇒ x2 – 1=3x
⇒ x2−3x– 1=0
Comparing equation x2−3x– 1=0 with general form ax2+bx+c=0,
We geta=1,b=−3 and c=−1
Using quadratic formula 
to solve equation,
⇒ 
⇒ 
(ii) 
⇒ 
⇒ 
⇒ −30=x2−7x+4x–28
⇒ x2−3x+2=0
Comparing equation x2−3x+2=0 with general form ax2+bx+c=0,
We get a=1,b=−3 and c=2
Using quadratic formula to solve equation,
⇒ 
⇒ 
⇒ x=2,1
5. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 13. Find his present age.
Ans. Let present age of Rehman= x years
Age of Rehman 3 years ago = (x−3) years.
Age of Rehman after 5 years = (x+5) years
According to the given condition:
⇒ 
⇒ 3(2x+2) =(x−3)(x+5)
⇒ 6x+6=x2−3x+5x−15
⇒ x2−4x– 15 – 6=0
⇒ x2−4x– 21=0
Comparing quadratic equation x2−4x– 21=0 with general form ax2+bx+c=0,
We get a=1,b=−4 and c=−21
Using quadratic formula
⇒ 
⇒ 
⇒ 
⇒ x=7,−3
We discard x=−3.Since age cannot be in negative.
Therefore, present age of Rehman is 7 years.
6. Two water taps together can fill a tank in
hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Ans. Let time taken by tap of smaller diameter to fill the tank = x hours
Let time taken by tap of larger diameter to fill the tank = (x– 10) hours
It means that tap of smaller diameter fills 
part of tank in 1 hour. … (1)
And, tap of larger diameter fills 
part of tank in 1 hour. … (2)
When two taps are used together, they fill tank in 758 hours.
In 1 hour, they fill 
part of tank 
… (3)
From (1), (2) and (3),
⇒ 
⇒ 75(2x−10)=8(x2−10x)
⇒ 150x– 750=8x2−80x
⇒ 8x2−80x−150x+750=0
⇒ 4x2−115x+375=0
Comparing equation 4x2−115x+375=0 with general equation ax2+bx+c=0,
We get a=4,b=−115andc=375
Applying quadratic formula
⇒ 
⇒ 
⇒ 
⇒ 
⇒ x=25,3.75
Time taken by larger tap = x– 10=3.75 – 10=−6.25 hours
Time cannot be in negative. Therefore, we ignore this value.
Time taken by larger tap = x– 10=25 – 10=15 hours
Therefore, time taken by larger tap is 15 hours and time taken by smaller tap is 25 hours.
7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.
(i) 2x2 – 3x + 5 = 0
(ii) 
(iii) 2x2 – 6x + 3 = 0
Ans. (i) 2x2 – 3x + 5 = 0
Comparing this equation with general equation ax2+bx+c=0,
We get a=2,b=−3 and c=5
Discriminant = b2−4ac=(−3)2 – 4(2)(5)=9 – 40=−31
Discriminant is less than 0 which means equation has no real roots.
(ii)
Comparing this equation with general equation ax2+bx+c=0,
We get a=3,b=
and c=4
Discriminant = b2−4ac=
−4(3)(4)=48 – 48=0
Discriminant is equal to zero which means equations has equal real roots.
Applying quadratic to find roots,
Because, equation has two equal roots, it means 
(iii) 2x2 – 6x + 3 = 0
Comparing equation with general equation ax2+bx+c=0,
We get a=2,b=−6, and c=3
Discriminant = b2−4ac=(−6)2 – 4(2)(3)=36 – 24=12
Value of discriminant is greater than zero.
Therefore, equation has distinct and real roots.
Applying quadratic formula to find roots,
⇒ 
⇒ 
8. If -4 is a root of the quadratic equation 
and the quadratic equation 
has equal root, find the value of k.
Ans. -4 is root of 
(Given)
[For equal roots D = 0]
9. Solve for 
Ans. 
Put 
10. 
solve for 
by factorization method.
Ans.
11. 
solve for by the method of completing the square.
Ans. 
12. Solve for : 
Ans. 
13. Using quadratic formula, solve for : 
Ans. 
14. In a cricket match, Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the numbers of wickets taken by these two is 15, find the number of wickets taken by each.
Ans. Let no. of wicket taken by Ravi =
No. of wicket taken by Kapil = 
According to question,
(Neglects)
So, no. of wickets taken by Ravi is 
15. The sum of a number and its reciprocal is 
. Find the number.
Ans. Let no. be
According to question,
