NCERT Solutions Class 10 Math Chapter 4 Quadratic Equations
NCERT Solutions Class 10 Math Chapter 4 Quadratic Equations: National Council of Educational Research and Training Class 10 Math Chapter 4 Solutions – Quadratic Equations. NCERT Solutions Class 10 Math Chapter 4 PDF Download.
NCERT Solutions Class 10 Math Chapter 4: Overview
Board |
NCERT |
Class |
10 |
Subject |
Math |
Chapter |
4 |
Chapter Name |
Quadratic Equations |
Topic |
Exercise Solutions |
NCERT Solutions Class 10 Math
Chapter 4 – Quadratic Equations
Exercise 4.1
(1) Check whether the following are quadratic equations:
(i) (x + 1)^{2} = 2 (x – 3)
Solution:
(x + 1)^{2} = 2 (x – 3)
= x^{2} + 2x + 1 = 2x – 6
= x^{2} + 7 = 0
It is not of the from ax^{2} + bx + c = 0
Hence, the given equation is not a quadratic equation.
(ii) x^{2} – 2x = (–2) (3 – x)
Solution:
x^{2} – 2x = (–2) (3 – x)
= x^{2} – 2x = -6 + 2x
= x^{2} – 4x + 6 = 0
It is in the form of ax^{2} + bx + c = 0
Hence, the given is a quadratic equation.
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
Solution:
(x – 2) (x + 1) = (x – 1) (x + 3)
= x^{2} – 2x – 2 + x = x^{2} – x + 3x – 3
= x^{2} – x – 2 = x^{2} + 2x – 3
= 3x – 1 = 0
It is not of the form ax^{2} + bx + c = 0
So, the given equation is not a quadratic equation.
(iv) (x – 3) (2x +1) = x(x + 5)
Solution:
(x – 3) (2x +1) = x(x + 5)
= 2x^{2} – 6x + x – 3 = x^{2} + 5x
= x^{2} – 11x + x – 3 = 0
= x^{2} – 10x – 3 = 0
It is in the form ax^{2} + bx + c = 0
Hence, the given equation is a quadratic equation.
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
Solution:
(2x – 1) (x – 3) = (x + 5) (x – 1)
= 2x^{2} – x – 6x + 3 = x^{2} + 5x – x – 5
= 2x^{2} – 7x + 3 = x^{2} + 4x – 5
= x^{2} – 11x + 8 = 0
It is in the form ax^{2} + bx + c = 0
Hence the given equation is a quadratic equation.
(vi) x^{2} + 3x + 1 = (x – 2)^{2}
Solution:
x^{2} + 3x + 1 = (x – 2)^{2}
= x^{2} + 3x + 1 = x^{2} – 4x + 4
= 7x – 3 = 0
It is not of the form ax^{2} + bx + c = 0
So, the given equation is not quadratic equation.
(vii) (x + 2)^{3} = 2x (x^{2} – 1)
Solution:
(x + 2)^{3} = 2x (x^{2} – 1)
= x^{3} + 6x^{2} + 12x + 8 = 2x^{3} – 2x
= – x^{3} + 6x^{2} + 14x + 8 = 0
It is not of the form ax^{2} + bx + c = 0
Hence, the given equation is not a quadratic equation.
(viii) x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}
Solution:
x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}
= x^{3} – 4x^{2} – x + 1 = x^{3} – 6x^{2} + 12x – 8
= 2x^{2 }– 13x + 9 = 0
It is of the form ax^{2} + bx + c = 0
So, the given equation is not a quadratic equation.
(2) Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m^{2}. 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.
Solution:
Let, the breadth = x m
So, the length = (2x + 1)
The area of the rectangular plot is
= breadth × length = x (2x + 1) m^{2} -*
According to question,
x (2x + 1) = 528
= 2x^{2 }+ x – 528 = 0
So, the length and breadth of the pilot satisfies the quadratic equation.
2x^{2} + x – 529 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let, one number = x
So, the another one = (x + 1)
Hence, x (x + 1) = 306
x^{2} + x – 306 = 0
So, the integers satisfies the quadratic equation,
x^{2} + x – 306 = 0
(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.
Solution:
Let, Rohan’s age = x year
So, his mother age = (x + 26) year
Then, 3 year from now rohan’s age = (x + 3) year and his mother age = (x + 26 + 3)
= (x + 29) year
Then, (x + 3) (x + 29) = 360
= x^{2} + 29x + 3x + 87 = 360
= x^{2} + 32x – 273 = 0
So, the rohan’s present age satisfies the above quadratic equation.
(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.
Solution:
Given, distance = 480 km
Let, speed = x km/h
So, time = distance/speed = 480/x h
If the speed had been 8km/h less then it would have taken 3 hours.
Now, 480/x = 480/x-8 + 3
= 480/x = 480+3(x-8)/(x-8)
= 480 (x – 8) = x {480 + 3 (x – 8)}
= 480x – 3840 = x (480 + 3x – 24)
= 480x – 3840 = 456x + 3x^{2}
= 3x^{2} – 24x + 3840 = 0
= x^{2} – 8x + 1280 = 0
So, the speed of the train satisfies the above quadratic equation.
EXERCISE 4.2
(1) Find the roots of the following quadratic equations by factorisation:
(i) x^{2} – 3x – 10 = 0
Solution:
x^{2} – 3x – 10 = 0
= x^{2} – (5 – 2) x – 10 =0
= x^{2} – 5x + 2x – 10 = 0
Or, x (x – 5) + 2 (x – 5) = 0
Or, (x + 2 (x – 5) = 0
x + 2 = 0
x = -2
x – 5 = 0
x = 5
Hence the roots of the equation are 5 or (-2).
(ii) 2x^{2} + x – 6 = 0
Solution:
2x^{2} + x – 6 = 0
Or, 2x^{2} + 4x – 3x – 6 = 0
Or, 2x (x + 2) – 3 (x + 2) = 0
Or, (2x – 3) (x + 2) = 0
2x – 3 = 0
2x = 3
x = 3/2
x + 2 = 0
x = -2
Hence, the roots of the equation are (-2) or 3/2.
(iii) √2 x^{2} + 7x + 5√2 = 0
Solution:
√2 x^{2} + 7x + 5√2 = 0
Or, √2 x^{2} + 5x + 2x + 5√2 = 0
Or, x (√2 x + 5) + √2 (√2 x + 5) = 0
Or, (x + √2) (√2x + 5) = 0
x + √2 = 0
x = -√2
√2x + 5 = 0
x = – 5/√2
Hence, the roots of the equation are (5/√2) or (-√2)
(iv) 2x^{2} – x + 1/8 = 0
Solution:
2x^{2} – x + 1/3 = 0
Or, 16x^{2} – 8x + 1 =
Or, 16x^{2} – 4x – 4x + 1 = 0
Or, 4x (4x – 1) – 1 (4x – 1) = 0
Or, (4x – 1) (4x – 1) = 0
Or, (4x – 1)^{2} = 0
4x – 1 = 0
Or, x = 1/4
4x – 1 = 0
x = 1/4
Hence, the roots of the equation are 1/4 or 1/4 .
(v) 100x^{2} – 20x + 1 = 0
Solution:
100x^{2} – 20x + 1 = 0
Or, 100x^{2} – 10x – 10x + 1 = 0
Or, 10x (10x – 1) – 1 (10x – 1) = 0
Or, (10x – 1) (10x – 1) = 0
10x – 1 = 0
x = 1/10
10x – 1 = 0
x = 1/1
Hence, roots of the equation are 1/10 or 1/10.
(2) Solve the problems given in Example 1.
Solution:
(i) Let the number of members john had be x.
Then the number of marbles Jivonti had = (45 – x)
The number of marbles left with John when he lost 5 marbles = (x – 5)
The number of marbles left with Jivonti when she lost 5 marbles = (45 – x – 5)
= (40 – x)
Therefore, their product = (x – 5) (40 – x)
= 40x – x^{2} – 200 + 5x
= -x^{2 }+ 45x – 200
So, – x^{2} + 45x – 200 = 124
= -x^{2} + 45x – 324 = 0
= x^{2} – 45x + 324 = 0
Therefore, the number of marbles john had, satisfies the quadratic equation.
x^{2} – 45x + 324 = 0
Now, x^{2} – 45x + 324 = 0
x^{2} – (36 + 9) x + 324 = 0
x^{2} – 36x – 9x + 324 = 0
x (x – 36) – 9 (x – 36) = 0
Or, (x – 36) (x – 9) = 0
Or, x – 36 = 0
x = 36
x – 9 = 0
x = 9
∴ If, the number of marbles John had = 36
Then the number of marbles Jivonti had = (45 – 36) = 9
If the number of marbles John had = 9
Then the number of marbles Jivonti had = 45 – 9 = 36
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production of each toy that day = (55 – x).
So, the total cost of production that day = x (55 – x)
Therefore, x (55 – x) = 750
Or, 55x – x^{2} = 750
Or, x^{2} – 55x + 750 = 0
Or, x^{2} – (30 + 25) x + 750 = 0
Or, x^{2} – 30x – 25x +750 = 0
Or, x (x – 25) – 25 (x – 30) = 0
Or, (x – 25) (x – 30) = 0
∴ x – 25 = 0
x = 25
x – 30 = 0
x = 3
Hence, the number of toys produced on that day be 25 or 30.
(3) Find two numbers whose sum is 27 and product is 182.
Solution:
Let, one number = x
So, other one (27 – x), As the sum of the numbers are 27.
Now, the product of the members = x (27 – x)
According to the question
x (27 – x) = 182
Or, 27x – x^{2} = 182
Or, x^{2} – 27x + 182 = 0
Or, x^{2} – (13 + 14) x + 182 = 0
Or, x^{2} – 13x – 14x + 182 = 0
Or, x (x – 13) – 14 (x – 13) = 0
Or, (x – 13) (x – 14) = 0
x – 13 = 0
x = 13
x – 14 = 0
x = 14
If one number is 13 then other = (27 – 13) = 14
If one number is 14 then other = (27 – 14) = 13
∴ Hence the numbers are = 13 and 14.
(4) Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Two consecutive positive integers, their sum of whose squares is 365.
Let, one number = x
∴ The other one = (x + 1)
According to the questions
x^{2} + (x + 1)^{2} = 365
Or, x^{2} + x^{2} + 2x + 1 = 365
Or, 2x^{2} + 2x – 364 = 0
Or, x^{2} + x – 182 = 0
Or, x^{2} + (14 – 13) x – 182 = 0
Or, x^{2} + 14x – 13x – 182 = 0
Or, x (x + 14) – 13 (x + 14) = 0
Or, (x + 14) (x – 13) = 0
x + 14 = 0
x = -14
x – 13 = 0
x = 13
Hence, the integers are positive, that’s why x = 13
∴ (x + 1) = 13 + 1 = 14 (other numbers)
Hence, the numbers are 13 and 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.
Solution:
Let the base = x cm
Height = (x – 7) cm
As per Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
= 13^{2} = (x – 7)^{2} + x^{2}
Or, 169 = x^{2} – 14x + 49 + x^{2}
Or, 2x^{2} – 14x – 120 = 0
Or, x^{2} – 7x – 60 = 0
Or, x^{2} – (12 – 5) x – 60 = 0
Or, x^{2} – 12x + 5x – 60 = 0
Or, x (x – 12) + 5 (x – 12) = 0
Or, (x – 12) (x + 5) = 0
Or, x – 12 = 0
x = 12
x + 5 = 0
x = -5
Length cannot be negative hence x ≠ -5
∴ Hence the base = 12cm and height = (12 – 7) cm = 5 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.
Solution:
It is given that total cost of production = 90 rupees
Let, No. of articles produced = x and cost of each article = (2x + 3) rupees
Total cost = No. of article × cost of each article.
Hence, x (2x + 3) = 90
= 2x^{2} + 3x – 90 = 0
= 2x^{2} + (15 – 12) x – 90 = 0
Or, 2x^{2} + 15x – 12x – 90 = 0
Or, x (2x + 15) – 6 (2x + 15) = 0
Or, (x – 6) (2x + 15) = 0
x – 6 = 0
x = 6
2x + 15 = 0
x = – 15/2
The number of article cannot be negative.
∴ Hence the number of articles produced is 6.
Cost of each article = (2x + 3) RS
= (2×6+3) RS
= 15 RS
EXERCISE 4.3
(1) Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x^{2} – 7x + 3 = 0
Solution:
2x^{2} – 7x + 3 = 0
Now, 2x^{2} – 7x + 3 = 0
Or, (√2 x)^{2} – 2. 7/2√2 . √2 x + (7/2√2)^{2 }– (7/2√2)^{2} + 3 = 0
Or, (√2 x – 7/2√2)^{2} – 49/8 + 3 = 0
Or, (√2 x – 7/2√2)^{2} + (-49+24/8) = 0
Or, (√2 x – 7/2√2)^{2} – 25/8 = 0
Or, (√2 x – 7/2√2)^{2} = 25/8
Or, (√2 x – 7/2√2) = ± 5/2√2
√2 x – 7/2√2 = 5/2√2
√2 x = 5/2√2 + 7/2√2
√2 x = 12/2√2
x = 12/2√2 × √2
= 12/4 = 3
√2x – 7/2√2 = – 5/2√2
Or, √2 x = 7/2√2 – 5/2√2
Or, √2 x = 2/2√2
Or, x = 1/√2 × √2 = 1/2
Hence the roots are 3 and 1/2.
(ii) 2x^{2} + x – 4 = 0
Solution:
2x^{2} + x – 4 = 0
Or, (√2 x)^{2} + 2. 1/2√2 . √2 x + (1/2√2)^{2} – (1/2√2)^{2} – 4 = 0
Or, (√2 x + 1/2√2)^{2} – 1/8 – 4 = 0
Or, (√2 x + 1/2√2)^{2} – (1 + 32/8) = 0
Or, (√2 x + 1/2√2)^{2} – 33/8 = 0
Or, (√2 x + 1/2√2)^{2} = 33/8
Or, √2 x + 1/2√2 = ± √33/2√2
√2 x + 1/2√2 = √33/2√2
Or, √2 x = √33/2√2 – 1/2√2
Or, √2 x = √33-1/2√2
Or, x = (√33 – 1)/4
√2 x + 1/2√2 = – √33/2√2
√2 x = – √33/2√2 – 1/2√2
√2 x = – √33-1/2√2
x = – (√33+1)/4
∴ Hence these are the two roots.
(iii) 4x^{2} + 4√3x + 3 = 0
Solution:
4x^{2} + 4√3x + 3 = 0
Or, (2x)^{2} + 2.2x √3 + (√3)^{2} – (√3)^{2} + 3 = 0
Or, (2x + √3)^{2} – 3 + 3 = 0
Or, (2x + √3)^{2} = 0
Or, (2x + √3) = 0
2x = – √3
x = – √3/2
Hence the root are (-√3/2) an (-√3/2)
(iv) 2x^{2} + x + 4 = 0
Solution:
2x^{2} + x + 4 = 0
Or, (√2 x)^{2} + 2. √2x/2√2 + (1/2√2)^{2} – (1/2√2)^{2} + 4 = 0
Or, (√2 x + 1/2√2)^{2} – 1/8 + 4 = 0
Or, (√2 x + 1/2√2)^{2} + (-1+32/8) = 0
Or, (√2 x + 1/2√2)^{2} + 31/8 = 0
Or, (√2 x + 1/2√2)^{2} = 31/8
But (√2 x + 1/2√2)^{2} cannot be negative for any real value of x. So there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.
(2) Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula
(i) 2x^{2} – 7x + 3 = 0
Here, a = 2, b = -7, c = 3
So, b^{2} – 4ac = 49 – 24 = 25 > 0
From the quadratic formula
x = -b± √b^{2} – 4ac/2a = 7± √25/4 = 7+5/4
Or, x = 7-5/4
x = 2/4 = 1/2
So, x = 7+5/4
= 12/4 = 3
∴ x = 3 or 1/2
(ii) 2x^{2} + x – 4 = 0
Here, a = 2, b = 1, c = -4
So, b^{2} – 4ac = 1 – 4.2 (-4) = 1+32 = 33 > 0
using x = -b± √b^{2} – 4ac/2a = -1± √33/4
Hence the roots are (-1+√33/4) an (-1-√33/4)
(iii) 4x^{2} + 4√3 x + 3 = 0
Here a = 4, b = 4√3, c = 3
So, b^{2} – 4ac = (4√3)^{2} – 4.4.3 = 48 – 48 = 0
Using x = -b± √b^{2} – 4ac/2a = -4√3 ±0/8 = – √3/2
So the roots is – √3/2
(iv) 2x^{2} + x + 4 = 0
Here a = 2, b = 1, c = 4
So, b^{2} – 4ac = 1 – 4.2.4 = 1 – 32 = 31 < 0
Since the square of a real number cannot be negative, therefore √b^{2} – 4ac will not have any real value.
So there are no real roots for the given equation.
(3) Find the roots of the following equations:
(i) x – 1/x = 3, x ≠ 0
Solution:
x – 1/x = 3
= x^{2} – 1 = 3x
Or, x^{2} – 3x – 1 = 0
Here, a = 1, b = -3, c = -1
So, b^{2} – 4ac = 9 – 4.1 (-1) = 9 + 4 = 13 > 0
Using, x = -b± √b^{2} – 40c/2a = 3± √13/2
∴ So the roots are 3 + √13/2 and 3 – √13/2
Exercise – 4.4
(1) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x^{2} – 3x + 5 = 0
Solution:
(i) 2x^{2} – 3x + 5 = 0
Here, a = 2, b = -3, c = 5
So, b^{2} – 4ac = (-3)^{2} – 4.2.5
= 9 – 40
= -31 < 0
Here, b^{2} – 4ac < 0
There is no real root exists for their equation.
(ii) 3x^{2} – 4√3 x + 4 = 0
Solution:
3x^{2} – 4√3 x + 4 = 0
Here, a = 3, b = -4√3, c = 4
b^{2} – 4ac = (-4√3)^{2} – 4.3.4 = 48 – 48 = 0
There are two equal real root.
x = -b±√b^{2} – 4ac/2.a
= +4√3/2×3
= 2√3/3
= 2/√3 (Ans)
(iii) 2x^{2} – 6x + 3 = 0
Solution:
2x^{2} – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
b^{2} – 4ac = (-6)^{2} – 2.2.3
= 36 – 24
= 12 > 0
There are two real root exist.
x = -b± √b^{2} – 4ac/2a
= 6±√12/4
= 2 (3 ± √3/4)
= 3±√3/2
The roots are x = 3+ √3/2 and 3 – √3/2
(2) Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x^{2} + kx + 3 = 0
Solution:
2x^{2} + kx + 3 = 0
Here, a = 2, b = k, c = 3
So, b^{2} – 4ac = k^{2} – 4.2.3 = k^{2} – 24
To have two equal roots, the discriminate should be zero.
k^{2} – 24 = 0
k^{2} = 24
Or, k = ± 2√6
For k = ± 2√6 the equation have two equal roots.
(ii) kx (x – 2) + 6 = 0
Solution:
kx (x – 2) + 6 = 0 k≠0
kx^{2} – 2kx + 6 = 0
Here, a = k, b = -2k, c = 6
So, b^{2} – 4ac
= (-2k)^{2} – 4.k.6
= 4k^{2} – 24k
To have two equal roots, the discriminant should be zero
4k^{2} – 24k = 0
Or, 4k (k – 6) = 0
Or, k (k – 6) = 0
K = 0, k = 6 but given k ≠ 0
For, k = 6 this equation have two equal roots.
(3) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^{2}? If so, find its length and breadth.
Solution:
Let, the breadth of the mango grove is x m. Then the length is 2 km.
The area is (2x × x) m^{2} = 2x^{2} m^{2}
Hence, 2x^{2} = 800
x^{2} – 400 = 0
Here, a = 1, b = 0, c = 400
So, the discriminant of the equation
b^{2} – 4ac = 0 – 4 × 2 × (-400) = 1600 > 0
∴ The roots are real
∴ The rectangular grove is possible the roots are –
x = -b± √b^{2} – 4ac/2a
= -0 ± √1600/2×0
= ± 40/2
= ±20
Length and breadth cannot be negative.
The breadth is = 20m and length is 40m
(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.
Solution:
Let, the age of first friend = x years
∴ The age of second friend = (20 – x) years
Four years ago the ages are (x – 4) and (20 – x – 4) = (16 – x) years respectively
According to the question
(x – 4) (16 – x) = 48
Or, 16x – x^{2} – 64 + 4x = 48
Or, x^{2} – 20x + 112 = 0
Here a = 1, b = -20, c = 112
So, the discriminant
b^{2} – 4ac = (-20)^{2} – 4.1.112
= 400 – 448
= -48 < 0
There will no real roots.
The situation is not possible.
(5) Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2}? If so, find its length and breadth.
Solution:
Let, the length of the park x m.
Then the breadth ½ (80 – 2x) m
We know that,
2 (Length + breadth) = Perimeter.
∴ The area of the park is 400m^{2}
So, x/2 (80 – 2x) = 400
Or, 40x – x^{2 }= 400
Or, x^{2} – 40x + 400 = 0
Here, a = 1, b = -40, c = 400
The discriminant
b^{2} – 4ac = (-40)^{2} – 4.1.400
= 1600 – 1600 = 0
There will be two equal real roots.
This rectangular park is possible with their parameter
The park is possible but it will be square is size become the length 20m.
Breadth = 1/2 (80 – 2×20) = 1/2 (80 – 40)
= 40/2 = 20 meter.