NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations

An equation of the form ax2+bx+c = 0, a ≠ 0 and a, b, c are real numbers is called quadratic equation.

(i) (x+1)2 = 2(x-3)
x2+2x+1 = 2x-6
x2+2x+1 -2x+6 = 0
x2+7 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1≠ 0, b = 0, c = 7
Hence the given equation is quadratic equation.

(ii) x2-2x = (-2)(3-x)
x2-2x  = -6+2x
x2-2x -2x+6 = 0
x2-4x+6 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1≠ 0, b = -4, c = 6
Hence the given equation is quadratic equation.

(iii) (x-2)(x+1) = (x-1)(x+3)
x2-2x+x-2 = x2-x+3x-3
x2-2x+x-2 -x2+x-3x+3= 0
-3x+1 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = -3, c = 1
Hence the given equation is not quadratic equation.

(iv) (x-3)(2x+1) = x(x+5)
x2-6x+x-3 = x2+5x
x2-6x+x-3 -x2-5x = 0
-10x-3 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = -10, c = -3
Hence the given equation is not quadratic equation.

(v) (2x-1)(x-3) = (x+5)(x-1)
2x2-6x-x+3 = x2+5x-x-5
2x2-6x-x+3 -x2-5x+x+5 = 0
x2-11x+8 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1 ≠ 0, b = -11, c = 8
Hence the given equation is quadratic equation.

(vi) x2+3x+1 = (x-2)2
x2+3x+1 = x2-4x+4
x2+3x+1 -x2+4x-4= 0
7x-3 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = 7, c = -3
Hence the given equation is not quadratic equation.

(vii) (x+2)3 = 2x(x2-1)
x3+6x2+12x+8 = 2x3-2x
x3+6x2+12x+8 -2x3+2x = 0
x3 – 6x2-14x-8 = 0
It is not in the form of ax2+bx+c = 0s
Hence the given equation is not a quadratic equation.

(viii) x3-4x2-x+1 = (x-2)3
x3-4x2-x+1 = x3-6x2+12x-8
x3-4x2-x+1 -x3+6x2-12x+8= 0
2x2-13x+9 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 2 ≠ 0, b = -13, c = 9
Hence the given equation is quadratic equation.

 

Represent the following situations in the form of quadratic equations:

 

(i)  Let the breadth of the plot is x meters
Then, the length is (2x + 1) meters.
Area = length × breadth
528 = x × (2x + 1)
528 = 2x2 + x
2x2 + x – 528 = 0

(ii)  Let x and x+1 be two consecutive integers.
Then, x(x + 1) = 306
x2 + x – 306 = 0

 

(iii)  Let x be the Rohan’s present age. Then, his mother’s present age is x+26
After 3 years theirs ages will be x+3 and x+26+3 = x+29 respectively.
Hence,
(x+3)(x+29) = 360
x2+3x+29x+87 = 360
x2+32x- 273 = 0

(iv) Let speed of train be x km/h

Time taken by train to cover 480 km = 480xhours

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)

This is a Quadratic Equation.

 

(i) x2-3x-10 = 0
x2-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
⇒ x-5 = 0 or, x+2 = 0
x = 5 or x =-2
Hence the required roots are 5, -2.

(ii) 2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = 3/2 or x = -2
Hence the required roots are 3/2 , -2.

(iii) √2x2+7x+5√2 = 0
√2x2+2x+5x+ 5√2 = 0
√2x(x+√2)+5(x+√2) = 0
(x+√2)(√2x +5) = 0
⇒ x+√2 = 0 or, √2x +5 = 0
x = -√2 or x = –5/√2

Hence the required roots are -√2, –5/√2.

 

(iv) 2x2-x+1/8=0
16x2-8x+1 = 0
(4x)2-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0  or 4x-1 = 0
x = 1/4   or x = 1/4
Hence the required roots are 1/4 , 1/4

(v)100x2-20x+1 = 0
(10x)2-2×10x.1+12= 0
(10x-1)2 = 0
10x-1 = 0 or 10x-1 = 0
x = 1/10 or x = 1/10
Hence the required roots are 1/10  or 1/10

 

(i)  Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x2+5x = 124
x2-45x+324 = 0
x2-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.

(ii)  Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x2 =750
x2-55x+750 = 0
x2-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.

 

Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x2-182 = 0
x2-27x+182 = 0
x2-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.

 

Let the two consecutive positive integers be x and x+1
Then by the given condition,
x2+(x+1)2 = 365
x2+ x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
x2+14x-13x-182 = 0
x(x+14)-13(x+14) = 0
(x+14)(x-13) = 0
x = -14 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.

Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x2+(x-7)2 = (13)2
x2+ x2-14x+49 = 169
2x2-14x-120 = 0
x2-7x-60 = 0
x2-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm

 

Let the number of articles produced on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x2+3x-90 = 0
2x2+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – 15/2
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = – 15/2 is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = 15

 

(i)2x2 – 7x + 3 = 0

First we divide equation by 2,

(ii) 2x2+ x – 4 = 0

Dividing equation by 2,

(iii) 4x2+4√3x+3 = 0

Dividing equation by 4,

(iv) 2x2+ x + 4 = 0

Dividing equation by 2,

 

(i) 2x2 – 7x + 3 = 0

Comparing quadratic equation 2x2 – 7x + 3 = 0 with general form ax2 + bx + c = 0, we get a = 2, b = -7 and c = 3

Putting these values in quadratic formula

(ii) 2x2+ x – 4 = 0

Comparing quadratic equation 2x2+ x – 4 = 0 with the general form ax2 + bx + c = 0 , we get a = 2, b = 1 and c = −4

Putting these values in quadratic formula ,

(iii) 4x2+4√3x+3 = 0

Comparing quadratic equation 4x2+4√3x+3 = 0 with the general form ax2 + bx + c = 0, we get b = and c = 3

Putting these values in quadratic formula ,

(iv) 2x2+ x + 4 = 0

Comparing quadratic equation 2x2+ x + 4 = 0 with the general form ax2 + bx + c = 0, we get a = 2, b = 1 and c = 4

Putting these values in quadratic formula

 

 

(i)

Comparing equation with general form ax2 + bx + c = 0,

We get a = 1, b = −3 and c = −1

Using quadratic formula

(ii)

 

⇒ x(x -2) -1 (x - 2) = 0

⇒ (x -1)(x - 2) = 0

⇒ x = 2, 1

 

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:

It is given that the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3.

⇒ 3 (2x + 2) = (x − 3) (x + 5)

⇒ 6x + 6 = x2 + 2x - 15

⇒x2 - 4x -21 = 0

⇒ x(x - 7) + 3(x - 7) = 0

⇒ (x - 7)(x + 3) = 0

⇒ x = 7, -3

However, age cannot be negative.

Therefore, Rehman's present age is 7 years.

 

Let Shefali’s marks in Mathematics = x

Let Shefali’s marks in English = 30 − x

If, she had got 2 marks more in Mathematics, her marks would be = x + 2

If, she had got 3 marks less in English, her marks in English would be = 30 – x − 3 = 27 − x

According to given condition:

(x + 2) (27 − x) = 210

⇒ - x2 + 25x + 156 = 0

⇒ x(x - 12) -13(x - 12) = 0

⇒ (x - 12)(x - 13) = 0

⇒ x = 12, 13

If the marks in Maths are 12, then marks in English will be 30 - 12 = 18

If the marks in Maths are 13, then marks in English will be 30 - 13 = 17

Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).

Let the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of rectangle =

It is given that the diagonal of the rectangle = (x+30)m

 

⇒ x2 + (x + 30)2 = (x + 60)2

⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x

⇒ x2 - 60x - 2700 = 0

⇒ x(x - 90) + 30(x -90)

⇒ (x - 90)(x + 30) = 0

⇒ x = 90, -30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.

Hence, length of the larger side will be (90 + 30) m = 120 m.

Therefore, length of sides are 90 and 120 in metres.

 

Let the larger and smaller number be x and y respectively.

According to the question,

x2 - y2 = 180 and y2 = 8x

⇒ x2 -8x = 180

⇒ x2 -8x - 180 = 0

⇒ x2 -18x + 10x - 180 = 0

⇒ x(x - 18) +10(x - 18) = 0

⇒ (x - 18)(x + 10) = 0

⇒ x = 18, -10

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

∴ y2 = 8x = 8 × 18 = 144

⇒ y =

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and - 12.

Therefore, two numbers are (12, 18) or (−12, 18)

 

 

Let the original speed of train be x km/hr. Then,

Increased speed of the train = (x + 5)km/hr

Time taken by the train under usual speed to cover 360 km =

Time taken by the train under increased speed to cover 360 km =

 

⇒ 1800 = x2 + 5x

⇒ x2 + 5x - 1800 = 0

⇒ x2 - 40x + 45x -1800 = 0

⇒ x(x - 40) + 45(x - 40) = 0

⇒(x - 40)(x + 45) = 0

So, either x - 40 = 0

x = 40

Or

x + 45 = 0

x = -45

But, the speed of the train can never be negative.

Hence, the original speed of train is x = 40 km/hr

 

Let the time taken by the smaller pipe to fill the tank be x hr.

Time taken by the larger pipe = (x - 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour =

It is given that the tank can be filled in  hours by both the pipes together. Therefore,

 

⇒75(2x - 10) = 8x2 - 80x

⇒ 150x -750 = 8x2 - 80x

⇒ 8x2 - 230x +750 = 0

⇒ 8x2 - 200x - 30x +750 = 0

⇒ 8x(x - 25) -30(x - 25) = 0

⇒ (x - 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 - 10 =15 hours respectively.

 

Let the speed of the passenger train be x km/hr. Then,

Speed of the express train = (x + 11)km/hr

Time taken by the passenger train to cover 132 km between Mysore to Bangalore = hours

Time taken by express train to cover 132 km =

According to the given condition,

⇒ 1452 = x2 + 11x

⇒ x2 - 33x + 44x -1452 = 0

x(x - 33) + 44(x - 33) = 0

(x - 33)(x + 44) = 0

So, either

x - 33 = 0

x = 33

Or

x + 44 = 0

x = -44

But, the speed of the train can never be negative.

Thus, when x = 33 then speed of express train

= x + 11

= 33 + 11

= 44

Hence, the speed of the passenger train is x = 33 km/hr

and the speed of the express train is x = 44 km/hr respectively.

 

Let the side of the first square be 'a' m and that of the second be ′A′ m.

Area of the first square = a2 sq m.

Area of the second square = A2 sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A - 4a = 24

A - a = 6          ......(1)

A2  + a2  = 468 ..........(2)

From (1), A = a + 6

Substituting for A in (2), we get

(a + 6)2 + a2 = 468

a2 + 12a + 36 + a2 = 468

2a2 + 12a + 36 = 468

a2 + 6a + 18 = 234

a2 + 6a - 216 = 0

a2 + 18a - 12a - 216 = 0

a(a + 18) - 12(a + 18) = 0

(a - 12)(a + 18) = 0

a = 12, - 18

So, the side of the first square is 12 m. and the side of the second square is 18 m.

 

 

Let the age of 1st friend is x.
Then the age of 2nd friend is (20-x)
Four year ago their age was (x-4) and (20-x-4)
Then using the given condition we have
(x-4)(16-x) = 48
16x – 64 – x2 +4x = 48
20x – x2 – 64 – 48 = 0
x2 – 20x + 112 = 0……(1)
Now comparing the above quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 1, b = -20 and c = 112
then, b2-4ac = (-20)2-4×1×112 = 400-448 = -48 > 0
Therefore, no real root is possible for this equation and hence this situation is not possible.

We know that the quadratic equation ax2+bx+c = 0 has
(a) Two distinct real roots, if b2-4ac > 0,
(b) Two equal real roots, if b2-4ac = 0,
(c) No real roots, if b2-4ac < 0.

 

(i) Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = -3 and c = 5
Then, b2-4ac = (-3)2-4×2×5 = 9-40 = -31 < 0
Hence the given quadratic equation has no real roots

 

(ii)  Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 3, b = -4√3 and c = 4
Then, b2-4ac = (-4√3)2-4×3×4 = 48-48 = 0
Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be

Therefore, the roots are

(ii)  Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = -6 and c = 3
then, b2-4ac = (-6)2-4×2×3 = 36-24 = 12 > 0
Hence the given quadratic equation has two distinct real roots.

Applying quadratic formulato find roots,

 

(i)  Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = k and c = 3
Now the given quadratic equation have two equal roots if
b2-4ac = 0

(k)2-4×2×3 = 0
k2-24 = 0
k2 = 24
k = ±√24
k = ± 2√6
Therefore, the required value of k is ± 2√6.

 

(ii)   The given quadratic equation can be written as
kx2-2kx+6 = 0….(1)
Comparing the quadratic equation (1) with the general form of quadratic equation ax2+bx+c = 0 we get
a = k, b = -2k and c = 6
Now the given quadratic equation have two equal roots if
b2-4ac = 0
(-2k)2-4×k×6 = 0
4k2-24k = 0
4k(k-6) = 0
k(k-6) = 0
k = 0
Or, k-6 = 0
k = 6
If k = 0, then the equation will not have x2 and x, which is not possible because the given equation is a quadratic equation.
Therefore, the required value of k is 6.

 

Let the breadth of the mango grove be x m and the length is 2x m.
Area = length × breadth
= x × 2x
= 2x2 m2
Then by the given condition,
2x2 = 800
x2 = 400
x = ±√400
x = ±20
Since, length cannot be negative, then x ≠ -20
Hence it is possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2 and its breadth is 20 m and length is 20×2 = 40 m

Let the length and breadth of the park be “l” and “b”
Area of rectangle = 2(l + b)
2(l + b) = 80
l + b = 40
b = 40 – l
Then, Area = l(40 – l)
Then by the given condition,
l(40 – l) = 400
40l  – l2 -400 = 0
l2 – 40l + 400 = 0….(1)
Now comparing the above quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 1, b = -40 and c = 400
Then, b2-4ac = (-40)2-4×1×400 = 1600-1600 = 0
As the quadratic equation has two equal roots then the given situation is possible.

Therefore, length of park, l = 20 m

And breadth of park, b = 40 - l = 40 - 20 = 20 m.