Quadratic Equations Class 10
Master Quadratic Equations Class 10 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Quadratic Equations Class 10 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 4.1
5 questionsEx 4.1, 1
Ex 4.1, 1 teackhoo
Check whether the following are quadratic equations :
(i) @e + 1)? = 2(x-3)
(x +1)? =2(x-3)
Using (a + b}? = a? + b’ + 2ab
x? 414+ 2x = 2x-6
x? 41+ 2x-2x +6=0
xv+1+6=0
w?+7=0
x7 +0x+7=0
Since, itis ofthe form ax? + bx +c =0
Wherea = 1,b = 0,c = 7
Hence, it is a quadratic equation
Ex 4.1, 2 (i)
Ex 4.1, 2 teackoo
Represent the following situations in the form of quadratic equations :
(i) The area of a 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.
= 2
Area = 528m Length = 2x +1
Also,
Length is one more that twice its breadth
Let Breadth = x
« Length = 2x +1
Ex 4.1, 2 (ii)
Ex 4.1, 2 teackhoo
Represent the following situations in the form of quadratic equations
(ii) The product of two consecutive positive integers is 306. We need
to find the integers.
There is a difference of 1 in consecutive integers,
Hence,
Let First integer = x
Second integer =x +1
Now, given that
Product of integers = 306
First integer x Second integer = 306
x (x + 1) = 306
Ex 4.1, 2 (iii)
Ex 4,1, 2 teachoo
Represent the following situations in the form of quadratic equations |
(iii} Rohan’s mother is 26 years older than him. The product of their
ages (in years) 3 years from now will be 360. We would like to find
Rohan’s present age.
Let Rohan's age = x
«. Rohan’s mother age = x + 26
After 3 years
Rohan’s age = x +3
Rehan’s mother age = (x + 26) + 3=x+29
Ex 4.1, 2 (iv)
Ex4.1,2 teachoo
Represent the following situations in the form of quadratic equations :
(iv) A train travels a distance of 480 km at a uniform speed. If the
speed had been 8 km/h less, then it would have taken 3 hours more
to cover the same distance. We need to find the speed of the train.
Let Speed of train be x km/hr
Normal speed Speed became 8 km/h less
Distance = 480 km Distance = 480 km
Speed = x km/hr Speed = (x— 8) km/hr
Time = = +3
Distance
Speed = Time
480 Distance
x= Time Speed ~ Time
480
Time = — _ 480
x x— 8 = 70a
+3)
480
(x-8) (*" +3) =480 (1)
Ex 4.2
10 questionsEx 4.2, 1 (i)
teackhoo
Ex 4,2,1
Find the roots of the following quadratic equations by factorization:
(i) x? -3x-10=0 Splitting the middle term
method
x2-3x-10=0 We need to find two numbers
whose
We factorize using
Sum = -—3
splitting the middle term method Product = -10x1= —-10
x? + 2x —5x-10=0 "lsum [Product |
x (x + 2)-5(x+2)=0 -10&1 -9 -10
(x5) (x +2) =0 -2&5 3 -10
2&-5 -3 -10
x-5=0 x+2=0
x=5 x=-2
Hence, x=5 and x= —2 are the roots of equation
Ex 4.2, 1 (ii)
Find the roots of the following quadratic equations by factorization:
(ii) 2x2 + x – 6 = 0
Ex 4.2, 1 (iii)
Find the roots of the following quadratic equations by factorization:
(iii) √2 x2 + 7x + 5 √2 = 0
Ex 4.2, 1 (iv)
Find the roots of the following quadratic equations by factorization:
(iv) 2x2 – x + 1/8=0
Ex 4.2, 1 (v)
Find the roots of the following quadratic equations by factorization:
(v) 100 x2 – 20x + 1 = 0
Ex 4.2, 2
Ex 4.2, 2 (i) teackhoo
Solve the problems given in Example 1 - x? - 45x + 324 =0
We found the equation as
x? — 45x + 324 =0
; Splitting the middle term
We factorize by method
splitting the middle term method We need to find two numbers
2 -9x-36x+324=0 | Whose
Sum = —45
x (x~ 9) ~ 36 (x—9)=0 Product = 324 x 1 = 324
(1-36) (x=) =0 Psu [Product |
81&4 85 324
x-36=0 x-9=0 -9&-36 -45 324
x=36 x=9
Thus, x = 36 & x =9 are the roots of equation
Ex 4.2, 3
teackhoo
Ex 4.2, 3
Find two numbers whose sum is 27 and product is 182.
Let First number be x
Given that Sum of both numbers is 27
First number + Second number = 27
x + Second number = 27
Second number = 27 - x
Also given that,
Product of both numbers = 182
First number X second number = 182
x (27 — x) = 182
27x —x? = 82
Ex 4.2, 4
Ex 4.2, 4 teachoo
Find two consecutive positive integers, sum of whose squares is 365.
There is difference of 1 in consecutive positive integers
Let First integer = x
-. Second integer = x+1
Given that
Sum of squares = 365
(First number)? + (Second number)? = 365
x? + (x +1)? = 365
x2 +x241742xxxX 1 = 365
2x? +14 2x = 365
Ex 4.2, 5
Ex 4.2, 5 teackhoo
The altitude of a right triangle is 7 cm less than its base. If the
hypotenuse is 13 cm, find the other two sides.
Let ABC be right angled triangle,
with altitude = AB, Base = BC & Hypotenuse = AC
Given
Hypotenuse = AC = 13 cm A
And, 13 cm
Altitude is 7 cm less than base x-7
B x Cc
Let Base = BC = x cm
-. Altitude = AB = Base—7 =x—7
Since ABC is a right angled triangle
Using Pythagoras theorem
Ex 4.2, 6
Ex 4.2, 6 teackhoo
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 Rs 90, find the
number of articles produced and the cost of each article.
We know that
Total Cost of Production
= Number of articles produced x Cost of article
Given
Total cost of production = Rs 90
Also,
Ex 4.3
8 questionsEx 4.3, 1 (i)
Ex 4.3, 1 teachoo
Find the nature of the roots of the following quadratic equations. If
the real roots exist, find them:
(i) 2x*-3x+5=0
2x? -3x+5=0
Comparing equation with ax? + bx +c =0
a=2,b= -3,c=5
We know that,
D=b2—4ac Concept:
D <0, no real roots
=(-3P -4x2x5 D = 0, two equal real roots
=9-—40 D > 0, two distinct real roots
= -31
Ex 4.3, 1 (ii)
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(ii) 3x2 – 4 √3 x + 4 = 0
Ex 4.3, 1 (iii)
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(iii) 2x2 – 6x + 3 = 0
Ex 4.3, 2 (i)
Ex 4.3, 2 teachoo
Find the values of k for each of the following quadratic equations, so
that they have two equal roots.
(i) 2x2 + kx + 3=0
2x? +kx+3=0
Comparing equation with ax? + bx +c=0
a=2,b=k,c=3
Since the equation has 2 equal roots,
D=0
b?- 4ac=0
Putting values
ke -4x2x3=0
k?-24=0
Ex 4.3, 2 (ii)
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(ii) kx (x – 2) + 6 = 0
Ex 4.3, 3
Ex 4.3, 3 teackhoo
Is it possible to design a rectangular mango grove whose length is
twice its breadth, and the area is 800 m?? If so, find its length and
breadth.
= 2
Area = 800 m Length = 2x
And,
Length is twice its breadth
Let Breadth be x
-. Length = 2x
Ex 4.3, 4
teachoo
Ex 4.3, 4
Is the following situation possible? If so, determine their present
ages. The sum of the ages of two friends is 20 years. Four years ago,
the product of their ages in years was 48.
Let the Present age of 1° friend =x
Given that
Sum of ages of friends is 20 years
Thus,
Present age of 1% friend + Present age of 2" friend = 20
x + Present age of 2" friend = 20
Present age of 2™ friend = 20 —x
Ex 4.3, 5
Ex 4.3, 5 teackhoo
Is it possible to design a rectangular park of perimeter 80 m and area
400 m2? If so, find its length and breadth.
Let Length be x
Given that
Perimeter = 80 m
2(Length + Breadth) = 80 m
80
(Length + Breadth) = 7
(x + Breadth) = 40
Breadth = 40 — x
Also, given that
Examples
22 questionsExample 1 (i)
Example 1 teackoo
Represent the following situations mathematically:
(i) John and Jivanti 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.
Total number of marbles = 45
Let the number of marbles with John = x
So, number of marbles with Jivanti = 45—x marbles
After losing 5 marbles,
Number of marbles with John = x—5
Number of marbles with Jivanti = 45 —x—5 = 40-x
Example 1 (ii)
Example 1 teachoo
Represent the following situations mathematically:
(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 Rs 750. We would like to find out the
number of toys produced on that day.
We know that
Total Cost of Production
= Number of toys produced x Cost of production of each toy
Given total cost of production = Rs 750
Let the number of toys produced per day = x
Example 2
Example 2 teackoo
Check whether the following are quadratic equations:
(i) (x- 2)2+1=2x-3
(x-2)?+1=2x-3
Using (a — b}? = a? + b?- 2ab
02 + 4— 4x) +1 = 2x-3
x2 4+5-—4x = 2x-3
+5—4x—2x+3=0
x? —-6x+8=0
It is the form ax? +bx+c=0
Where, a=1,b= -—6,c=8
Hence, it is a quadratic equation .
Example 3
Example 3 teackoo
Find the roots of the equation 2x? — 5x + 3 = 0, by factorization.
2x?-5x+3=0
Splitting the middle term method
We factorize by
We need to find two numbers whose
splitting the middle term method
Sum=—-5
2x? - 3x -2x+3=0
Product =3x2=6
(2x -— 3) (x-1)=0 126 7 6
-3&-2 -5 6
2x-3=0 x-1=0
2x =3 x=1
=3
X=>
Hence, x = > and X = 1 are the roots of the equation
Example 4
Example 4 teackoo
Find the roots of the quadratic equation 6x? — x— 2 = 0.
2 _ _ =
Gxt x—2=0 Splitting the middle term method
We factorize by We need to find two numbers
splitting the middle term method whose
> Sum=-1
6x! — 4x + 3x-2=0 Product = -2x6= —12
NODS | sum [Product |
(2x + 1) (3x-2)=0 -2&6 4 -12
-4&3 -1 -12
2x+1=0 3x-2=0
2x=-1 3x =2
-1 2
x=— x=-
2 3
Hence, x= > and x= ; are the roots of the equation
Example 5
Example 5 teachoo
Find the roots of the quadratic equation 3x2 - 2 Véx+2=0.
Splitting the middle term
3x°-2V6x+2=0 method
We will factorize We need to find two numbers
using splitting the middle term method| whose
3x2 -V6x—V6x+2=0 Sum = - 2v6
Product = 2 x 3=6
3x? -V¥2x 3x —-—V2x 3x+2=0 Isum [Produc |
(V2) (3)x — (23) Je a—Vé -2V6 6
V3 x(v3x — V2) —V2 (v3x-—v2) =0
(V3x — V2) (V/3x—+v2) =0
V¥3x —v2=0 | V3x —v2=0
V¥3x=v2 V3x = v2
2_ ip eeze P
X= B43 “Ve 3
Example 6
Example 6 teackoo
Find the dimensions of the prayer hall discussed in Section 4.1.
Suppose a charity trust decides to build a prayer hall having a carpet
area of 300 square metres with its length one metre more than twice
its breadth. What should be the length and breadth of the hall?
Area of hall = Length x breadth
Length = 2x+1
Given,
Area of hall = 300 m2
Also,
length is one metre more than twice its breadth
Let Breadth of the hall = x metres
-. Length of hall = (2x + 1) metres.
Example 7
Example 7 teachoo
Find the discriminant of the quadratic equation 2x? — 4x + 3 = 0, and
hence find the nature of its roots.
2x? -4x+3=0
Comparing equation with ax? + bx + c=0
a=2,b=-4,c=3
We know that
D =b?-4ac
D = (-4)? - 4x (2) x (3)
D=16-24
D=-8
Example 8
Example 8 teackoo
A pole has to be erected at a point on the boundary of a circular park
of diameter 13 metres in sucha way that the differences of its
distances from two diametrically opposite fixed gates A and B on the
boundary is 7 metres. Is it possible to do so? If yes, at what distances
from the two gates should the pole be erected?
<8
Let P be the pole
Gates A & B are diametrically opposite )
AB = Diameter of circle = 13 m
Also given that,
Difference of the distance of the pole from the two gates is 7 metres
BP — AP or AP— BP is 7m
Let us take AP — BP =7
Example 9
Example 9 teackoo
Find the discriminant of the equation 3x2 — 2x + : = 0 and hence
find the nature of its roots. Find them, if they are real.
3x? — 2x +2 =0
3x 3x°- 3x 2x41 =0
3
9x?-6x+1=0x3
9x? -6x+1=0
Comparing equation with ax? + bx +c=0
a=9,b=-6,c=1
We know that
Question 1
Example 7 teachoo.com
Solve the equation given in Example 3 (2x? — 5x + 3 = 0) by the
method of completing the square.
2x*-5x+3=0
Dividing by 2
ax?-Sx+3_ 0
2 ~ 2
2x? 5x | 3
272 t27?
25% 13 _
x 2 + 3 0
We know that
{a—b)* = a? - 2ab + b?
Here, a=x&
—~2ab = =
2
— 5x
— 2xb = = — (As a =x)
Question 2
Example 8 teachoo.com
Find the roots of the equation 5x? — 6x — 2 = 0 by the method of
completing the square.
5x°-6x-2=0
Dividing by 5
5x?-6x-2_ 0
5 5
Sai 6x 2 _ 9
5 5 5
x2 — 6x 2 =0
5 5
We know that
(a—b)* = a? — 2ab + b?
Here, a=x &
—2ab = 76x
5
- 2xb === (As a =x)
Question 3
Example 9 teachoo.com
Find the roots of 4x? + 3x + 5 = 0 by the method of completing the
square.
4x? + 3x+5=0
Dividing by 4
axt 3x 5g
4 44
e+ 4S 9
44
We know that
(a+ b)* = a? + 2ab + b?
Here, a=x &
2ab ==
4
2xb = 2% (As a=x)
4
2b=2
4
Question 4
teachoo.com
Example 10
Solve Q. 2 (i) of Exercise 4.1 by using the quadratic formula.
The equation we formed in Ex4.1,2 (i) was
2x? +x— 528 =0
Comparing equation with ax? + bx + c =0
Here a=2, b=1,c= —528
We know that
D =b?-4ac
D=(1)? —4x 2 x (—528)
D=1-8x (— 528)
D=1+8x528
D=14+ 4224
D=4225
Question 5
Example 11 teachoo.com
Find two consecutive odd positive integers, sum of whose squares is
290.
There is a difference of 2 in odd consecutive integers,
Hence,
Let the first integer = x
Second integer =x + 2
Also, given that
Sum of the squares of both the numbers = 290
(First number)? + (Second number)? = 290
(x)? + (x + 2P = 290
Using (a + b}? = a? + b? + 2ab
x? +x24+4+4 4x = 290
2x? + 4x +4-290=0
Question 6
Example 12 teachoo.com
A rectangular park is to be designed whose breadth is 3 m less than
its length. Its area is to be 4 square metres more than the area of a
park that has already been made in the shape of an isosceles
triangle with its base as the breadth of the rectangular park and of
altitude 12 m (see Fig.}. Find its length and breadth.
Rectangle Triangle
Let breadth =
et brea x D C
Since breadth is 3 mless__| Area of triangle 43
than length, 4
. ==x Base x Altitude
length is 3 m more than 2
breadth 1 ; A * B
==xCD x Altitude
So, Length =x +3 2
uty AB x Altitude (Opposite sides of
2 rectangle are equal,
Area = Length x Breadth
1
= x(x + 3) Hp%x x 12
= 6X
Question 7 (i)
Example 13 teachoo.com
Find the roots of the following quadratic equations, if they exist,
using the quadratic formula:
(i) 3x*-5x+2=0
3x? -5x+2=0
Comparing equation with ax? + bx + c=0
Here, a=3,b= —-5,c=2
We know that,
D=b?-4ac
D=(-5)? - 4x (3) x @)
D=25-—24
D=1
So, the roots of the equation is given by
2a
Question 7 (ii)
Example 13
Find the roots of the following quadratic equations, if they exist, using the quadratic formula:
(ii) x2 + 4x + 5 = 0
Question 7 (iii)
Example 13
Find the roots of the following quadratic equations, if they exist, using the quadratic formula:
(iii) 2x2 2 2 + 1 = 0
Question 7 (i)
Example 14 teachoo.com
Find the roots of the following equations:
(i) xt2=3,x40
x
x+2=3
x
x(x)t1 =3
x
xed =3
x
x? +1 = 3x
x?-3x+1=0
We will factorize by quadratic formula
Comparing equation with ax? + bx + c=0
Here, a=1,b=-3,c=1
Question 7 (ii)
Example 14
Find the roots of the following equations:
(ii) 1/𝑥−1/(𝑥−2)=3,𝑥≠0,2
Question 8
Example 15 teachoo.com
A motor boat whose speed is 18 km/h in still water takes 1 hour more
to go 24 km upstream than to return downstream to the same spot.
Find the speed of the stream.
Given that speed of the boat = 18 km/ hr.
Let the speed of the stream =x km / hr.
For upstream For downstream
mN
Speed of boat
Speed of boat
Speed of stred . Speed of strea'
ae
Speed = (18 — x) km/hr Speed = (18 + x) km/hr
Distance = 24 km Distance = 24 km
Distance _ Distance
Speed = Time upstream Speed = Time downstream
24
2 = —__*
(18 _ x) = —_*4__ (18 * x) Time downstream
Time upstream 24
a 24 Time downstream = ———
Time upstream = —_—— (18 + x)
(18 - x)
Case Based Questions (MCQ)
2 questionsQuestion 1
Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.
Question 1
What will be the distance covered by Ajay’s car in two hours?
(a) 2(x + 5) km
(b) (x - 5) km
(c) 2(x + 10) km
(d) 2(x + 5) km
Question 2
Which of the following quadratic equation describe the speed of Raj’s car?
(a) x
2
-5x-500=0
(b) x
2
- 4x-400=0
(c) x
2
+5x-500=0
(d) x
2
-4x+400=0
Question 3
What is the speed of Raj’s car?
(a) 20 km/hour
(b) 15 km/hour
(c) 25 km/hour
(d) 10 km/hour
Question 4
How much time took Ajay to travel 400 km?
(a) 20 hour
(b) 40 hour
(c) 25 hour
(d) 16 hour
Question 2
The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.
Question 1
Let speed of the stream be
x
km/hr. then speed of the motorboat in upstream will be
(a) 20 km/hr
(b) (20 + x) km/hr
(c) (20 – x) km/hr
(d) 2 km/hr
Question 2
What is the relation between speed ,distance and time?
(a) speed = (distance )/time
(b) distance = (speed )/time
(c) time = speed x distance
(d) speed = distance x time
Question 3
Which is the correct quadratic equation for the speed of the current ?
(a) x
2
+30x-200=0
(b) x
2
+20x-400=0
(c) x
2
+30x-400=0
(d) x
2
-20x-400=0
Question 4
What is the speed of current ?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour
Question 5
How much time boat took in downstream?
(a) 90 minute
(b) 15 minute
(c) 30 minute
(d) 45 minute
Completing the square and Word Problems
18 questionsQuestion 1 (i)
Ex 4.3 ,1 teachoo.com
Find the roots of the following quadratic equations, if they exist,
by the method of completing the square:
(i) 2x*-7x+3=0
2x? — 7x +3 =0
Dividing by 2
ax? -7x+3=0 0
2 ~ 2
2? _ 7x 3 _
2 2°52
2 7% 3 _
x 2 + 2 0
We know that
(a—b)* = a? - 2ab + b?
Here, a=x&
-Jab= -2
2
-2xb=-Z (Asa=x)
Question 1 (ii)
Ex 4.3 ,1 teachoo.com
Find the roots of the following quadratic equations, if they exist, by
the method of completing the square:
(ii) 2x2+x-4=0
2x? +x-4=0
Dividing whole equation by 2
274-4 0
2 ~ 2
2x x 4
227277?
x42 2=0
2
We know that
{a +b)? = a? + 2ab + b?
Here, a=x&
2ab ==
2
2xb = 5 (As a=x)
Question 1 (iii)
Ex 4.3 ,1 teachoo.com
Find the roots of the following quadratic equations, if they exist,
by the method of completing the square:
(iii) 4x2 + 4V3x +3 =0
4x2 +4¥3x+3=0
Dividing whole equation 4
4x24 4 V¥3x+3 _ 0
4 ~ 4
2
Axt + 4v3 + 3_ 0
4 4 4
43x + == 0
We know that
(a + b}? = a2 + 2ab + b?
Here, a=x&
2ab = 3x
2xb = ¥3x (As a =x)
Question 1 (iv)
Ex 4.3 ,1 teachoo.com
Find the roots of the following quadratic equations, if they exist, by
the method of completing the square:
(iv) 2x? +x+4=0
2X°+x+4=0
Dividing equation by 2
2xr?+xt4 0
2 ~ 2
2x7 x 4
Tz + 3 + 3 =0
24% =
x +5 +2=0
We know that
(a + b)* = a? + 2ab + b?
Here, a=x &
2ab ==
2
2xb =~ (As a =x)
9
Question 2 (i)
Ex 4.3 ,2 teachoo.com
Find the roots of the quadratic equation using quadratic formula
(i) 2x*-7x+3=0
2x? -7x+3=0
Comparing equation with ax? + bx +c =0
a=2,b=-7,c=3
We know that
D =b?- 4ac
D=(-77 -4x2x3
D=(-7 x -7)-(4xX2x3)
D=49-24
D=25
Question 2 (ii)
Find the roots of the quadratic equation using quadratic formula
(ii) 2x2 + x 4 = 0
Question 2 (iii)
Find the roots of the quadratic equation using quadratic formula
(iii) 4x2 + 4 3 +3=0
Question 2 (iv)
Find the roots of the quadratic equation using quadratic formula
(iv) 2x2 + x + 4 = 0
Question 3 (i)
Ex 4.3 ,3 teachoo.com
Find the roots of the following equations:
(i) x-2 =3,x #0
x
x-2=3
x
x(x} -1 =3
x
wr1i3
x
x*-1= 3x
x*- 3x-1=0
Comparing equation with ax? + bx + c=0
So, a=1,b= -3,c= -1
We know that
D =b?- 4ac
D=(-3)? -4xK1x-1
Question 3 (ii)
Find the roots of the following equations:
(ii) 1/(𝑥 + 4)−1/(𝑥 − 7)=11/30,𝑥≠−4, 7
Question 4
teachoo.
Ex 4.3 ,4 enemee eos
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago
and 5 years from now is :: Find his present age.
Let Rehman's current age = x
Rehman’s age 3 years ago = x— 3
Rehman’s age 5 years from now =x+5
Given that
Sum of reciprocal of Rehman’s ages 3 yr. ago and 5 yr. from now =5
a
(Age of Rehman3 yrago) (Age of rehmans yr from now) “3
1 1 1
x-3 + KtS 3
X+54x-3 a
(«-3)« +5) 3
2x+2 _ a
(«-3)« +5) 3
Question 5
Ex 4.3 ,5 teachoo.com
In a class test, the sum of Shefali’s marks in Mathematics and English
is 30. Had she got 2 marks more in Mathematics and 3 marks less in
English, the product of their marks would have been 210. Find her
marks in the two subjects.
Let shefali’s marks in mathematics = x
Given that sum of Shefali’s marks in English and Mathematics is 30
-. Marks in mathematics + Marks in English = 30
x + Marks in English = 30
Marks in English = 30 —x
Also, it is given that
had she got 2 marks more in mathematics and 3 marks less in
English, the product of their marks would have been 210
-. (Marks in Maths + 2) X (Marks in English — 3) = 210
(x + 2) (30-—x—3}= 210
Question 6
EX 4.3 ,6 teachoo.com
The diagonal of a rectangular field is 60 metres more than the
shorter side. If the longer side is 30 metres more than the shorter
side, find the sides of the field.
A D
Let ABCD be the rectangular field
xm
AC is diagonal
B x+30m Cc
Let the shorter side = AB = x metres
It is given that,
Diagonal is 60 m more than shorter side
AC =AB+60=x+ 60 metres
Also longer side is 30 m more than shorter side
BC = AB + 30=x+ 30 metres
Question 7
Ex 4.3 ,7 teachoo.com
The difference of squares of two numbers is 180. The square of the
smaller number is 8 times the larger number. Find the two numbers.
Let larger number be x
Given,
The square of the smaller number is 8 times the larger number
(Smaller number)? = 8 x Larger number
(Smaller number)? = 8x
Smaller number = +V¥8x
Also, given that
Difference of squares of two numbers is 180
-. Square of larger number — square of smaller number = 180
(Larger number)? — (Smaller number}? = 180
Question 8
Ex 4.3 ,8 teachoo.com
A train travels 360 km at a uniform speed. If the speed had been 5
km/h more, it would have taken 1 hour less for the same journey. Find
the speed of the train.
Let the speed of train be x km/hr
Normal speed Speed 5 km/h more
Distance = 360 km Distance = 360 km
Speed = x km/hr Speed = {x +5) km/hr
Time = 2-14
x
Speed = Distance
Time
Distance
= 360 Speed = Time
Time
360
. 360 +5-—
Time = = x (2-1)
360
(+5) Ce - 1) =480. ...(1)
Question 9
Ex 4.3 ,9 teachoo.com
Two water taps together can fill a tank in 9 : 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.
Let the time taken by smaller tap to fill tank completely = x hours
So, Volume of tank filled by smaller tap in 1 hour ==
Also, it is given that
time take by larger tap is 10 hour less
Time taken by larger tap to fill tank completely
= Time taken by smaller tap — 10
=x-—10 hours
Volume of Tank filled by larger tap in 1 hour = —
Question 10
Ex 4.3 ,10 teachoo.com
An express train takes 1 hour less than a passenger train to travel 132
km between Mysore and Bangalore (without taking into
consideration the time they stop at intermediate stations). If the
average speed of the express train is 11km/h more than that of the
passenger train, find the average speed of the two trains.
Let the average speed of passenger train =x km/h
Passenger train Express train
Distance = 132 km Distance = 132 km
Speed =x Given that speed of express train is
11 km/h more than passenger
Distance So, Speed =x +11
Speed = ——_
Time passenger
y=e—_ 2 Speed = Distance _
Time passenger Time express
. 132 x+11= 13? _
Time passenger = a Time express
Time express = —=2—
P ~ (+11)
Question 11
Ex 4.3 ,11 teachoo.com
Sum of the areas of two squares is 468 m2. If the difference of their
perimeters is 24 m, find the sides of the two squares.
Let side of square 1 be x metres
Perimeter of square 1 = 4 x Side = 4x
Now, it is given that
Difference of perimeter of squares is 24 m
Perimeter of square 1— Perimeter of square 2 = 24
4x — perimeter of square 2 = 24
Ax — 24 = perimeter of square 2
Perimeter of square 2 = 4x — 24
Now,
Perimeter of square 2 = 4x — 24
4x (Side of square 2 ) = 4x— 24
Side of square 2 = Mf 7 24 “a-9) = y
Why Learn This With Teachoo?
Quadratic Equations is Chapter 4 of NCERT Class 10 Mathematics. It covers identification and formation of quadratic equations, solution by factorisation, completing the square and the quadratic formula, nature of roots and word problems. Teachoo provides detailed solutions for Exercises 4.1 to 4.3, examples, case-based questions and special practice on completing the square.
What is a quadratic equation?
A quadratic equation in x can be written as ax² + bx + c = 0, where a, b and c are real and a ≠ 0. An equation may need simplification before its degree is clear. The roots are values of x that satisfy the equation.
Students form quadratic equations from geometrical, numerical and practical situations. Because quadratic models may produce two algebraic roots, each answer must be checked against contextual restrictions such as positive length or time.
Methods of solving
Factorisation rewrites the quadratic as a product of two linear factors and applies the zero-product property. Splitting the middle term is effective when suitable integer or rational factors exist.
Completing the square rewrites the equation in the form (x + p)² = q and reveals the structure behind the general formula. The quadratic formula is
x = [−b ± √(b² − 4ac)]/(2a).
Coefficients must be taken from standard form, including their signs.
Nature of roots
The discriminant D = b² − 4ac determines the nature of real roots:
-
D > 0: two distinct real roots;
-
D = 0: two equal real roots;
-
D < 0: no real roots.
When D is a positive perfect square and coefficients are rational, roots are rational; when positive but not a perfect square, roots are irrational.
Topics available on Teachoo
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Exercises 4.1 to 4.3 and examples;
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checking and forming quadratic equations;
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splitting the middle term;
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word problems using factorisation;
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completing the square;
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quadratic formula;
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nature of roots; and
-
case-based questions.
Learning outcomes
Students should be able to identify and form quadratics, solve them by all three methods and use the discriminant without finding roots when appropriate. They should verify roots by substitution and reject solutions that violate the original context.
Why is this chapter important?
Quadratics model area, motion, number relationships and optimisation. The chapter is high-value for board exams because it combines algebraic accuracy, formula use and application modelling.
How Teachoo helps
Teachoo groups questions by solution method. First place the equation in standard form and identify a, b and c with signs. Try factorisation if the structure is friendly; otherwise use completing the square or the formula. Keep the ± symbol until both roots are obtained.
For word problems, define the variable, form the equation, solve and interpret. Compare with Teachoo’s steps to identify whether an error occurred in modelling, algebra or final selection.
Board-exam and competency preparation
Quadratic questions commonly ask for roots, nature of roots, a parameter value or a real-life dimension. Standard form is the compulsory first step because a, b and c cannot be read safely from an unsimplified equation. If the question asks only for the nature of roots, use the discriminant and avoid unnecessary full solutions.
In case-based problems, one root may violate a physical condition even though it satisfies the algebraic equation. State why it is rejected. When a parameter appears, translate “equal roots,” “distinct real roots” or “no real roots” into a condition on D. For accuracy, calculate the discriminant on a separate line and simplify the radical before writing both roots.
Quick revision checklist
Solve one quadratic by each method, classify roots from three discriminants, form an equation from a word problem and solve a parameter question. Substitute roots back and confirm that accepted answers satisfy contextual restrictions.
Common mistakes to avoid
Do not use a = 0 in a quadratic model. In the formula, the denominator is 2a for the entire numerator. Calculate b² before subtracting 4ac and preserve signs. A negative discriminant means no real roots, not no roots in every extended number system.
Deeper reasoning and concept connections
Study Quadratic Equations through comparison and justification. Place two related examples side by side, identify the decisive difference and explain why one method works in each case. Then create a new example and a deliberate non-example. This forces the definition to do real work and exposes gaps that passive reading hides.
Students should also practise reversing questions. After solving for an answer, ask what question could have produced it, whether more than one answer is possible and which extra condition would make the result unique. Reverse reasoning develops flexibility and is especially useful for missing-value, assertion–reason and error-analysis questions. The goal is to understand the network of ideas, not merely the order of a textbook solution.
How to solve unfamiliar and competency-based questions
Begin by separating facts from conclusions. Facts are given by the question or a known property; conclusions must be derived. Draw or rewrite the problem so each fact has a visible place. If several methods are possible, prefer the one with fewer assumptions and an easy final check. Record intermediate results rather than doing everything mentally.
Competency questions often change context without changing mathematics. Replace names and story details with variables, shapes, sets or data values. After solving, restore the context and check feasibility: counts should be whole where required, lengths and areas should have suitable units, probabilities should lie between 0 and 1, and constructed figures should satisfy every stated condition.
What complete mastery looks like
For Quadratic Equations, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Quadratic Equations?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Quadratic Equations?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
What makes an equation quadratic?
After simplification, its highest variable power is two and the coefficient of x² is non-zero.
What does the discriminant tell us?
It determines whether the equation has two distinct real roots, equal real roots or no real roots.
Must both roots of a word problem be accepted?
No. Each must satisfy the original equation and the contextual conditions.
Standardise, solve, verify and interpret. Skipping any one of these steps creates avoidable errors.