NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations - PDF Download

The NCERT Solutions for Class 10 Maths Chapter 4 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.

Download Solution PDF

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

Below are the NCERT class 10 maths chapter 4 solutions for exercise questions.

Q1 (i): Check whether the following are quadratic equations : (x+1)² = 2(x−3)

Answer:

We have L.H.S.

(x+1)² = x² + 2x + 1

Therefore, (x+1)² = 2(x−3) can be written as:

⇒ x²+2x+1 = 2x−6

i.e., x²+7=0

Or x²+0x+7=0
This equation is of the type: ax²+bx+c=0

Hence, the given equation is a quadratic equation.

Q1 (ii): Check whether the following are quadratic equations: x²−2x=(−2)(3−x)

Answer:

Given equation x²−2x=(−2)(3−x) can be written as:

⇒x²−2x=−6+2x

i.e., x²−4x+6=0

This equation is of the type: ax²+bx+c=0

Hence, the given equation is a quadratic equation.

Q1 (iii): Check whether the following are quadratic equations : (x−2)(x+1)=(x−1)(x+3)

Answer:

L.H.S.

(x−2)(x+1) can be written as:

= x²+x−2x−2 = x²−x−2

and R.H.S (x−1)(x+3) can be written as:

= x²+3x−x−3=x²+2x−3 ⇒ x²−x−2 = x²+2x−3

i.e., 3x−1=0

The equation is not of the type: ax²+bx+c = 0, a≠0.

Hence, the given equation is not a quadratic equation since a=0.

Q1 (iv): Check whether the following are quadratic equations : (x−3)(2x+1)=x(x+5)

Answer:

L.H.S.

(x−3)(2x+1) can be written as:

= 2x²+x−6x−3 = 2x²−5x−3

and R.H.S (x)(x+5) can be written as:

= x²+5x ⇒ 2x²−5x−3 = x²+5x

i.e., x²−10x−3 = 0

This equation is of type: ax²+bx+c = 0, a≠0.

Hence, the given equation is a quadratic equation.

Q1 (v): Check whether the following are quadratic equations : (2x−1)(x−3)=(x+5)(x−1)

Answer:

L.H.S. (2x−1)(x−3) can be written as:

= 2x²−6x−x+3 = 2x²−7x+3

and R.H.S (x+5)(x−1) can be written as:

= x²−x+5x−5 = x²+4x−5 ⇒ 2x²−7x+3 = x²+4x−5

i.e., x²−11x+8 = 0

This equation is of type: ax²+bx+c=0, a≠0.

Hence, the given equation is a quadratic equation.

Q1 (vi): Check whether the following are quadratic equations: x²+3x+1=(x−2)²

Answer:

L.H.S. x²+3x+1
and R.H.S (x−2)² can be written as:

= x²−4x+4 ⇒ x²+3x+1 = x²−4x+4

i.e., 7x−3=0

This equation is not of the type: ax2+bx+c=0, a≠0

Here, a=0, hence, the given equation is not a quadratic equation.

Q1 (vii): Check whether the following are quadratic equations : (x+2)³=2x(x²−1)

Answer:

L.H.S.

(x+2)³ can be written as:

= x³+8+6x(x+2)=x³+6x²+12x+8

and R.H.S

2x(x²−1) can be written as:

= 2x³−2x ⇒ x³+6x²+12x+8=2x³−2x

i.e., x³−6x²−14x−8=0

This equation is NOT of type: ax²+bx+c=0.

Hence, the given equation is not a quadratic equation.

Q1 (viii): Check whether the following are quadratic equations: x³−4x²−x+1=(x−2)³

Answer:

L.H.S. x³−4x²−x+1, and R.H.S. (x−2)³ can be written as:

= x³−6x²+12x−8 ⇒ x³−4x²−x+1 = x³−6x²+12x−8

i.e., 2x²−13x+9=0

This equation is of the type: ax²+bx+c=0.

Hence, the given equation is a quadratic equation.

Q2 (i): Represent the following situations in the form of quadratic equations: The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Answer:

Given that the area of a rectangular plot is 528 m2.
Let the breadth of the plot be ' b ' .
Then, the length of the plot will be: =2b+1.
Therefore, the area will be:
= b(2b+1) m², which is equal to the given plot area, 528 m²
⇒ 2b²+b = 528
⇒ 2b²+b−528 = 0
Hence, the length and breadth of the plot will satisfy the equation 2b²+b−528=0

Q2 (ii): Represent the following situations in the form of quadratic equations: The product of two consecutive positive integers is 306. We need to find the integers.

Answer:

Given that the product of two consecutive integers is 306.
Let two consecutive integers be ' x ' and ′x+1′.
Then, their product will be:
x(x+1)=306
⇒ x²+x−306=0
Hence, the two consecutive integers will satisfy this quadratic equation x² + x − 306 = 0.

Q2 (iii): Represent the following situations in the form of quadratic equations: 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.

Answer:

Let the age of Rohan be 'x′ years.
Then his mother's age will be: 'x+26′ years.
After three years,
Rohan's age will be 'x+3′ years, and his mother's age will be ' x+29′ years.
Then, according to the question,
The product of their ages 3 years from now will be:
⇒ (x+3)(x+29)=360
⇒ x²+3x+29x+87=360
⇒ x²+32x−273=0
Hence, the age of Rohan satisfies the quadratic equation x²+32x−273=0.

Q2 (iv): Represent the following situations in the form of quadratic equations: 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.

Answer:

Let the speed of the train be 's' km/h.

The distance to be covered by the train is 480 km.

The time taken will be

= 480/s hours

Now, according to the question

$\frac{480}{s-8}-\frac{480}{s}=3$

⇒ 480s−480(s−8)=3s(s-8)

⇒ 3840=3s²−24s

⇒ 3s²−24s−3840=0

Dividing by 3 on both sides

s²−8s−1280=0

Hence, the speed of the train satisfies the quadratic equation s²−8s−1280=0

Q1 (i): Find the roots of the following quadratic equations by factorization: x²−3x−10=0

Answer:

Given the quadratic equation: x²−3x−10=0

Factorization gives x²−5x+2x−10=0

⇒ x²−5x+2x−10=0

⇒ x(x−5)+2(x−5)=0

⇒ (x−5)(x+2)=0

⇒ x=5 or −2
Hence, the roots of the given quadratic equation are 5 and -2.

Q1 (ii): Find the roots of the following quadratic equations by factorization: 2x²+x−6=0

Answer:

Given the quadratic equation: 2x2+x−6=0

Factorisation gives 2x²+4x−3x−6=0

⇒ 2x(x+2)−3(x+2)=0

⇒ (x+2)(2x−3)=0

⇒ x=−2 or 32

Hence, the roots of the given quadratic equation are −2 and 32.

Q1 (iii): Find the roots of the following quadratic equations by factorization: 2x²+7x+52=0

Answer:

Given the quadratic equation: 2x²+7x+52=0

Factorization gives 2x²+5x+2x+52=0

⇒ x(2x+5)+2(2x+5)=0

⇒ (2x+5)(x+2)=0

⇒ x=−52 or −2
Hence, the roots of the given quadratic equation are −52 and −2.

Q1 (iv): Find the roots of the following quadratic equations by factorization: 2x2−x+18=0

Answer:

Given the quadratic equation: 2x²−x+18=0

Solving the quadratic equations, we get

16x²−8x+1=0

Factorization gives

⇒ 16x²−4x−4x+1=0

⇒ 4x(4x−1)−1(4x−1)=0

⇒ (4x−1)(4x−1)=0

⇒ x=1/4 or 1/4

Hence, the roots of the given quadratic equation are 1/4 and 1/4.

Q1 (v): Find the roots of the following quadratic equations by factorization: 100x²−20x+1=0

Answer:

Given the quadratic equation: 100x²−20x+1=0

Factorisation gives 100x²−10x−10x+1=0

⇒ 10x(10x−1)−1(10x−1)=0

⇒ (10x−1)(10x−1)=0

⇒ x = $\frac1{10}$

Hence, the roots of the given quadratic equation are $\frac1{10}$.

Q2: Solve the problems given in Example 1. (i) x²−45x+324=0
_(ii) x²−55x+750=0

Answer:

From Example 1, we get:

Equations:

(i) x²−45x+324=0
Solving by the factorization method:

Given the quadratic equation: x²−45x+324=0

Factorization gives x²−36x−9x+324=0

⇒ x(x−36)−9(x−36)=0

⇒ (x−9)(x−36)=0

⇒ x=9 or 36
Hence, the roots of the given quadratic equation are x = 9 and 36.

Therefore, John and Jivanti have 36 and 9 marbles, respectively, in the beginning.

(ii) x²−55x+750=0

Solving by the factorization method:
Given the quadratic equation: x²−55x+750=0
Factorization gives x²−30x−25x+750=0

⇒ x(x−30)−25(x−30)=0

⇒ (x−25)(x−30)=0

⇒ x = 25 or 30
Hence, the roots of the given quadratic equation are x= 25 and 30.
Therefore, the number of toys on that day was 30 or 25.

Q3: Find two numbers whose sum is 27 and whose product is 182.

Answer:

Let two numbers be x and y.

Then, their sum will be equal to 27, and the product equals 182.

x+y=27……(1)

xy=182……(2)

From equation (2) we have:

y=182/x
Then, putting the value of y in equation (1), we get x+182/x = 27

Solving this equation:

⇒ x²−27x+182=0

⇒ x²−13x−14x+182=0

⇒ x(x−13)−14(x−13)=0

⇒ (x−14)(x−13)=0

⇒ x=13 or 14
Hence, the two required numbers are 13 and 14.

Q4: Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

Let the two consecutive integers be 'x' and 'x+1′.
Then the sum of the squares is 365.

x²+(x+1)²=365

⇒ x²+x2+1+2x=365

⇒ x²+x−182=0

⇒ x²−13x+14x+182=0

⇒ x(x−13)+14(x−13)=0

⇒ (x−13)(x−14)=0

⇒ x=13 or 14
Hence, the two consecutive integers are 13 and 14.

Q5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:

Let the length of the base of the triangle be b cm.

Then, the altitude length will be: b-7 cm.

Given if the hypotenuse is 13 cm.

Applying the Pythagoras theorem, we get

Hypotenuse 2= Perpendicular 2+ Base 2

So, (13)²=(b−7)²+b²

⇒ 169=2b2+49−14b

⇒ 2b²−14b−120=0

⇒ b²−7b−60=0

⇒ b²−12b+5b−60=0

⇒ b(b−12)+5(b−12)=0

⇒ (b−12)(b+5)=0

⇒ b=12 or −5
But the length of the base cannot be negative.
Hence, the base length will be 12 cm.
Therefore, we have Altitude length =12 cm - 7 cm = 5 cm and Base length = 12 cm.

Q6: 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.

Answer:

Let the number of articles produced in a day = x

The cost of production of each article will be = 2x + 3

Given that the total production on that day was Rs.90.

Hence, we have the equation;

x(2x+3)=90

⇒ x²+3x−90=0

⇒ 2x²+15x−12x−90=0

⇒ x(2x+15)−6(2x+15)=0

⇒ (2x+15)(x−6)=0

⇒ x=−152 or 6
But, x cannot be negative as it is the number of articles.
Therefore, x=6 and the cost of each article = 2x+3 = 2(6)+3 = 15

Hence, the number of articles is 6 and the cost of each article is Rs 15.

Q1 (i): Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

2x²−3x+5=0

Answer:

For a quadratic equation, ax2+bx+c=0, the value of the discriminant determines the nature of the roots and is equal to:
D = b²−4ac

If D>0, then the roots are distinct and real.
If D<0, then no real roots.
If D=0, then there exist two equal real roots.
Given the quadratic equation, 2x2−3x+5=0.
Comparing with the general to get the values of a,b, and c.

a=2,b=−3,c=5
Finding the discriminant:

D = (−3)²−4(2)(5) = 9−40 = −31
∵D<0

Here, D is negative,
Hence, there are no real roots possible for the given equation.

Q1 (ii): Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 3x²−4$\sqrt3$x+4=0

Answer:

b²−4ac=(−4$\sqrt3$)²−(4×4×3)=48−48=0

Here, the value of the discriminant =0, which implies that roots exist and the roots are equal.
The roots are given by the formula
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt3\pm0}{2×3}=\frac{2}{\sqrt3}$
So the roots are $\frac{2}{\sqrt3}$, $\frac{2}{\sqrt3}$

Q1 (iii): Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

2x²−6x+3=0

Answer:

The value of the discriminant

b²−4ac = (−6)²−4×2×3 = 12

The discriminant >0.

Therefore, the given quadratic equation has two distinct real roots.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm2\sqrt3}{2×2}=\frac{3}{2}\pm\frac{\sqrt3}{2}$

So the roots are $\frac{3}{2}\pm\frac{\sqrt3}{2}$.

Q2 (i): Find the values of k for each of the following quadratic equations so that they have two equal roots.

2x²+kx+3=0

Answer:

For two equal roots for the quadratic equation: ax²+bx+c=0
The value of the discriminant D = 0.
Given equation: $2x^2+kx+3=0$

Comparing and getting the values of a,b, and c.
a=2,b=k,c=3
The value of $D=b^2-4ac=k^2-4(2)(3)=k^2-24=0$
⇒ $k=\pm\sqrt{24}=\pm2\sqrt{6}$

Q2 (ii): Find the values of k for each of the following quadratic equations so that they have two equal roots

kx(x−2)+6=0

Answer:

For two equal roots of the quadratic equation: ax² + bx + c = 0

The value of the discriminant is: $D=0$

Given equation: $kx(x-2)+6=0$

Rewriting it in standard quadratic form: $kx^2-2kx+6=0$

Comparing with $ax^2+bx+c=0$ we get:

$a=k$, $b=-2k$, and $c=6$

The value of the discriminant:$\begin{array}{r}\end{array}$

$D=b^2-4ac=4k^2-4(k)(6)=0$
⇒ $k = 0$ or $k=6$

But $k = 0$ is not possible because it will not satisfy the given equation.

Hence, the only possible value of $k$ is 6 to get two equal roots.

Q3: Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Answer:

Let the breadth of the mango grove be 'b'.

Then the length of the mango grove will be '2b'.

And the area will be:

Area = (2b)(b) =2b²
Which will be equal to 800 m2 according to question.

⇒ 2b²=800 m²

⇒ b²−400 = 0
Comparing to get the values of a,b,c.

⇒ a=1,b=0,c=−400
Finding the discriminant value:

D = b²−4ac = 0²−4(1)(−400) =1600
Here, D>0

Therefore, the equation will have real roots.

And hence finding the dimensions:

⇒b²−400=0

⇒b = ±20
As a negative value is not possible, the value of the breadth of the mango grove will be 20 m.
And the length of the mango grove will be = 2×20 =40 m

Q4: 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.

Answer:

Let the age of one friend be x years.

And the age of another friend will be: (20−x) years.
4 years ago, their ages were x−4 years and 20−x−4 years.
According to the question, the product of their ages in years was 48.

∴(x−4)(20−x−4)=48

⇒16x−64−x²+4x=48

⇒−x²+20x−112=0

⇒x²−20x+112=0
Now, comparing to get the values of a,b,c.

a=1,b=−20,c=112
Discriminant value D=b2−4ac=(−20)2−4(1)(112)=400−448=−48 As D<0.

This situation is not possible.

Q5: Is it possible to design a rectangular park of perimeter 80 m and area 400 m ² ? If so, find its length and breadth.

Answer:
Let us assume the length and breadth of the park be ' l ' and ' b ' respectively.
Then, the perimeter will be

P = 2(l+b) = 80

⇒ l+b = 40 or b = 40−l
The area of the park is:

Area = l×b = l(40−l) = 40l−l2
Given: 40l−l2 = 400
l²−40l+400 = 0

Comparing to get the values of a,b and c.
The value of the discriminant $D=b^2-4ac=(-40)^2-4(1)(400)=0$
As D=0.
Therefore, this equation will have two equal roots.

And hence the roots will be:

l=40/2(1)=40/2=20

Therefore, the length of the park, l = 20 m and the breadth of the park b = 40 − l = 40 − 20 = 20 m.

Quadratic Equations Class 10 Solutions - Exercise Wise

Here are the exercise-wise links for the NCERT class 10 chapter 4, Quadratic Equations:

• NCERT Class 10 Maths Chapter 4 Exercise 4.1
• NCERT Class 10 Maths Chapter 4 Exercise 4.2
• NCERT Class 10 Maths Chapter 4 Exercise 4.3

Quadratic Equations Class 10 Chapter 4: Topics

Topics you will learn in NCERT Class 10 Maths Chapter 4 Quadratic Equations include:

• 4.1 Introduction
• 4.2 Quadratic Equations
• 4.3 Solution of a Quadratic Equation by Factorisation
• 4.4 Nature of Roots

Class 10 Maths Chapter 4 Quadratic Equations - Important Formulae

Quadratic Equation

A quadratic equation is a polynomial with degree two in the form $ax^2+bx+c=0$ where a, b, c are real numbers and a≠0.

A real number α is said to be a root of the quadratic equation if α satisfies the given quadratic equation.

Quadratic Formula

The quadratic formula to find the roots of the quadratic equation is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where b²−4ac≠0.
Here, $b^2−4ac$ is called the discriminant.
Sum of roots = $-\frac{b}{a}$

Product of roots = $\frac{c}{a}$

Quadratic Equation in terms of roots

The general form of a quadratic equation when the roots α and β are given is x² − (α+β)x + (αβ) = 0

Nature of Roots of the Quadratic Equation

The nature of the roots of the quadratic equation depends on the value of the discriminant(D), i.e. b²−4ac

• D > 0: Roots are real and distinct.

• D = 0: Roots are real and equal.

• D < 0: Roots are imaginary.

NCERT Solutions for Class 10 Maths: Chapter Wise

We at Careers360 compiled all the NCERT class 10 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Exemplar Solutions Subject-wise

Given below are the links for subject-wise exemplar solutions. Go through the solutions for a better understanding of the concepts.

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 10:

• NCERT Books Class 10 Maths
• NCERT Syllabus Class 10 Maths
• NCERT Books Class 10
• NCERT Syllabus Class 10