NCERT solution for class 10 maths quadratic equations ( Chapter 4)

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Solution for Exercise 4.1

Q1. Check whether the following are Quadratic Equations.
(i)\((x + 1)^2 = 2(x - 3)\)
(ii)\(x^2 - 2x = (-2) (3 - x)\)
(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)\(x^2 + 3x + 1 = (x - 2)^2\)
(vii)\((x + 2)^3 = 2x(x^2 - 1)\)
(viii)\(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

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Answer :

(i)\((x + 1)^2 = 2(x - 3)\)
Using identity \((a + b)^2 = a^2 + 2ab + b^2\)
=>\(x^2 + 1 + 2x = 2x - 6\)
=>\(x^2 + 7 = 0\)
Hence it is a quadratic equation since the degree is 2.

(ii)\(x^2 - 2x = (-2) (3 - x)\)
=>\(x^2 - 2x = -6 + 2x\)
=>\(x^2 - 2x - 2x + 6 = 0\)
=>\(x^2 - 4x + 6 = 0\)
Hence it is a quadratic equation with degree 2.

(iii)\((x - 2) (x + 1) = (x - 1) (x + 3)\)
=>\(x^2 + x - 2x - 2 = x^2 + 3x - x - 3\)
=>\(x^2 + x - 2x - 2 - (x^2 + 3x - x - 3) = 0\)
=>\(x^2 - x^2 + x - 2x - 3x + x - 2 + 3 = 0\)
=>\(-3x + 1 = 0\)
Hence it is a not quadratic equation since the degree is 1.

(iv)\((x - 3) (2x + 1) = x (x + 5)\)
=>\(2x^2 + x - 6x - 3 = x^2 + 5x\)
=>\(2x^2 + x - 6x - 3 - x^2 - 5x = 0\)
=>\(x^2 - 10x - 3 = 0\)
Hence it is a quadratic equation since the degree is 2.

(v)\((2x - 1) (x - 3) = (x + 5) (x - 1)\)
=>\(2x^2 - x - 6x + 3 = x^2 - x + 5x - 5\)
=>\(2x^2 - 7x + 3 - x^2 + x - 5x + 5 = 0\)
=>\(x^2 - 10x - 3 = 0\)
Hence it is a quadratic equation since the degree is 2.

(vi)\(x^2 + 3x + 1 = (x - 2)^2\)
Using identity \((a - b)^2 = a^2 - 2ab + b^2\)
=>\(x^2 + 3x + 1 = x^2 - 4x + 4 \)
=>\(x^2 + 3x + 1 - x^2 + 4x - 4 = 0\)
=>\(7x - 3 = 0\)
Hence it is a not quadratic equation since the degree is 1.

(vii)\((x + 2)^3 = 2x(x^2 - 1)\)
Using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
=>\(x^3 + 2^3 + 3(x)(2)(x + 2) = 2x(x^2 - 1)\)
=>\(x^3 + 8 + 6x(x + 2) = 2x^3 - 2x\)
=>\(2x^3 - 2x - x^3 - 8 - 6x^2 - 12x = 0\)
=>\(x^3 - 6x^2 - 14x - 8 = 0\)
Hence it is a not quadratic equation since the degree is 3

(viii)\(x^3 - 4x^2 - x + 1 = (x - 2)^3\)
Using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
=>\(x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3(x)(2)(x-2)\)
=>\(-4x^2 - x + 1 = -8 - 6x^2 + 12x\)
=>\(2x^2 - 13x + 9 = 0\)
Hence it is a quadratic equation since the degree is 2

Q.2 Represent the following situations in the form of Quadratic Equations:
(i) The area of 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
(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.

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Answer :

(i) A rectangular plot with area 528 \(m^2\) is given.
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
=> \(528 = x (2x + 1)\)
=> \(528 = 2x^2 + x\)
=>\( 2x^2 + x – 528 = 0\)
This is the Quadratic Equation.

(ii) Let two consecutive numbers be x and (x + 1).
It is given that \(x (x + 1) = 306\)
=>\(x^2 + x = 306\)
=>\( x^2 + x – 306 = 0\)
This is the 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:
\((x + 3) (x + 29) = 360\)
=>\(x^2 + 29x + 3x + 87 = 360\)
=>\(x^2 + 32x - 273 = 0\)
This is the Quadratic Equation.

(iv) Let speed of train be x km/h
Time taken by train to cover 480 km = \(\frac{480}{x}\) hours
If, speed had been 8km/h less then time taken would be \(\frac{480}{x - 8}\) hours.
According to given condition, if speed had been 8km/h less then time taken is 3 hours less.
Therefore, \(\frac{480}{x – 8} = \frac{480}{x} + 3\)
=>\( 480 (\frac{1}{x – 8} - \frac{1}{x}) = 3\)
=>\( 480 (x – x + 8) = 3 (x) (x - 8) \)
=>\( 480 × 8 = 3 (x) (x - 8)\)
=>\( 3840 = 3x^2 - 24x\)
=>\(3x^2 - 24x - 3840 = 0\)
Dividing equation by 3, we get
=>\(x^2 - 8x - 1280 = 0\)
This is the Quadratic Equation.

Solution for Exercise 4.2

Q1. Find the roots of the following Quadratic Equations by factorization.
(i) \(x^2 - 3x - 10 = 0\)
(ii) \(2x^2 + x - 6 = 0\)
(iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
(iv) \(2x^2 - x + {{1} \over 8} = 0\)
(v) \(100x^2 - 20x + 1 = 0\)

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Answer :

(i) \(x^2 -3x - 10 = 0\)
(Splitting -3x as 2x - 5x)
\( x^2 + 2x - 5x - 10 = 0 \)
\( x(x + 2) - 5(x + 2) = 0 \)
\( (x - 5)(x + 2) = 0 \)
The roots of this equation are the values of x for which \( (x - 5)(x + 2) = 0 \)
which are,
\( x - 5 = 0 \) or \( x + 2 = 0\)
Thus, \(x = 5\) or \(x = -2\)

(ii)
\(2x^2 +x - 6 = 0\)
(Splitting x as 4x - 3x)
\(2x^2 +4x - 3x - 6 = 0\)
\( 2x(x + 2) - 3(x + 2) = 0 \)
\( (2x - 3)(x + 2) = 0 \)
The roots of this equation are the values of x for which \( (2x - 3)(x + 2) = 0 \)
which are,
\( 2x - 3 = 0 \) or \( x + 2 = 0\)
Thus, \(x = \frac{3}{2}\) or \(x = -2\)

(iii)
\(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
(Splitting 7x as 2x + 5x)
\(\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \)
\( \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \)
\( (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \)
The roots of this equation are the values of x for which \( (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \)
which are,
\(\sqrt{2}x + 5 = 0 \) or \( x + \sqrt{2} = 0\)
Thus, \(x = \frac{-5}{\sqrt{2}}\) or \(x = \sqrt{-2}\)

(iv)
\(2x^2 – x + \frac { 1 }{ 8 } = 0 \)
Multiplying and dividing the entire equation with 8, we get -
\(\frac{1}{8}(16x^2 - 8x + 1) = 0\)
(Splitting -8x as -4x - 4x)
\(\frac{1}{8}(16x^2 - 4x - 4x + 1) = 0\)
\( \frac{1}{8}[4x(4x - 1) - 1(4x - 1)] = 0 \)
\( \frac{1}{8}(4x - 1)(4x - 1) = 0 \)
\( \frac{1}{8}(4x - 1)^2= 0 \)
The roots of this equation are the values of x for which \( \frac{1}{8}(4x - 1)^2= 0 \)
which is,
\( 4x - 1 = 0\)
Thus, \(x = \frac{1}{4}\)

(v)
\(100 x^2 – 20x + 1 = 0\)
(Splitting -20x as - 10x - 10x)
\(100 x^2 – 10x - 10x + 1 = 0\)
\( 10x(10x - 1) - 1(10x - 1) = 0 \)
\( (10x - 1)(10x - 1) = 0 \)
\( (10x - 1)^2 = 0 \)
The roots of this equation are the values of x for which \( (10x - 1)^2 = 0 \)
which is,
\(10x - 1 = 0 \)
Thus, \(x = \frac{1}{10}\)

Q2.Solve 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.
(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 ?750. We would like to find out the number of toys produced on that day.


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Answer :

(i) Let the number of marbles John has be \(x\).
Thus, the number of marbles Jivanti has will be \(45 - x\).
After losing 5 marbles each, number of marbles left with John will be \(x - 5\)
and with Jivanti will be \(45 - x - 5 \)
= \(40 - x \)
Since the product of their final number of marbles is given to be 124,
\((x - 5)(40 - x) = 124\)
\(40x - x^2 - 200 + 5x = 124\)
\(-x^2 + 45x - 324 = 0\)
\(x^2 - 45x + 324 = 0\)
(Splitting -45x as -9x - 36x )
\(x^2 - 9x - 36x + 324 = 0\)
\(x(x - 9) - 36(x - 9) = 0\)
\((x - 36)(x - 9) = 0\)
The roots of this equation are the values of x for which \( (x - 36)(x - 9) = 0 \)
which are,
\( x - 36 = 0 \) or \( x - 9 = 0\)
Thus, \(x = 36\) or \(x = 9\)

So, if the number of marbles with John is 36, the number of marbles with Jivanti will be 45 - 36 = 9.
If the number of marbles with John is 9, the number of marbles with Jivanti will be 45 - 9 = 36.


(ii) Let the number of toys produced in a day be \(x\)
Thus, the cost of production of each toy will be \(55 - x\) Rs.
Since the total production cost is given to be 750 Rs,
\(x(55 - x) = 750\)
\(55x - x^2 - 750 = 0\)
\(x^2 - 55x + 750 = 0\)
(Splitting -55x as -30x - 25x )
\(x^2 - 30x - 25x + 750 = 0\)
\(x(x - 30) - 25(x - 30) = 0\)
\((x - 25)(x - 30) = 0\)
The roots of this equation are the values of x for which \( (x - 25)(x - 30) = 0 \)
which are,
\( x - 25 = 0 \) or \( x - 30 = 0\)
Thus, \(x = 25\) or \(x = 30\)

So, if the number of toys produced in a day is 25, the cost of production of each toy will be 55 - 25 = 30Rs.
If the number of toys produced in a day is 30, the cost of production of each toy will be 55 - 30 = 25Rs.
Thus, the number of toys produced on that day are either 25 or 30.

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

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Answer :

Let the first number be \(x\)
Thus the second number can be represented as \(27 - x\).
Since the product of the two numbers is 182,
\(x(27 - x) = 182\)
\(27x - x^2 - 182 = 0\)
\(x^2 - 27x + 182 = 0\)
(Splitting -27x as -13x - 14x )
\(x^2 - 13x - 14x + 182 = 0\)
\(x(x - 13) - 14(x - 13) = 0\)
\((x - 14)(x - 13) = 0\)
The roots of this equation are the values of x for which \( (x - 14)(x - 13) = 0 \)
which are,
\( x - 14 = 0 \) or \( x - 13 = 0\)
Thus, \(x = 14\) or \(x = 13\)

If the first number is 14, the second number will be 27 - 14 = 13.
If the first number is 13, the second number will be 27 -13 = 14.
Thus, the two numbers are 13 and 14.

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

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Answer :

Let the first number be \(x\)
The second number can be represented by \(x + 1\).
From the given condition,
\(x^2 + (x + 1)^2 = 365\)
\(x^2 + x^2 + 2x + 1 - 365 = 0\)
\(2x^2 + 2x -364 = 0\)
\(x^2 + x - 182 = 0\)
(Splitting x as 14x - 13x )
\(x^2 + 14x - 13x - 182 = 0\)
\(x(x + 14) - 13(x + 14) = 0\)
\((x - 13)(x + 14) = 0\)
The roots of this equation are the values of x for which \( (x + 14)(x - 13) = 0 \)
which are,
\( x + 14 = 0 \) or \( x - 13 = 0\)
Thus, \(x = -14\) or \(x = 13\)

Since the integer is supposed to be positive, \(x = -14\) is eliminated.
Thus, the two consecutive positive integers are 13 and 13 + 1 = 14.

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

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Answer :

Let the base of the right triangle be \(x\).
The altitude can be represented by \(x - 7\).
In a right triangle,
\(Hypotenuse^2 = Base^2 + Altitude^2\) (Pythagoras Theorem)
\(13^2 = x^2 + (x - 7)^2 \)
\(169 = x^2 + x^2 - 14x + 49 \)
\(2x^2 - 14x - 120 = 0 \)
\(x^2 - 7x - 60 = 0\)
(Splitting -7x as 5x - 12x )
\(x^2 + 5x - 12x - 60 = 0\)
\(x(x + 5) - 12(x + 5) = 0\)
\((x - 12)(x + 5) = 0\)
The roots of this equation are the values of x for which \( (x - 12)(x + 5) = 0 \)
which are,
\( x + 5 = 0 \) or \( x - 12 = 0\)
Thus, \(x = -5\) or \(x = 12\)

Since the length of any side cannot be negative, we reject \(x = -5\).
Thus, the length of the base is 12 cm and the length of the altitude is 12 - 7 = 5 cm.

Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that 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.

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Answer :

Let the number of articles produced be \(x\).
Thus, the cost of production of each article can be expressed as \(2x + 3\).
As the total cost of production on that day is 90 Rs.,
\(x(2x + 3) = 90\)
\(2x^2 + 3x - 90 = 0\)
(Splitting 3x as -12x + 15x )
\(2x^2 + -12x +15x - 90 = 0\)
\(2x(x - 6) + 15(x - 6) = 0\)
\((2x + 15)(x - 6) = 0\)
The roots of this equation are the values of x for which \((2x + 15)(x - 6) = 0 \)
which are,
\(2x + 15 = 0 \) or \(x - 6 = 0\)
Thus, \(x = \frac{-15}{2}\) or \(x = 6\)

Since the number of articles produced cannot be a negative number, \(x = \frac{-15}{2}\) is rejected.
Thus, the number of articles produced is 6 and the cost of each article is \(2*6 + 3\) = 15Rs.

Solution for Exercise 4.3

Q1. Find the roots of the following quadratic equations if they exist by the method of completing square.
(i) \(2x^2 - 7x + 3 = 0\)
(ii) \(2x^2 + x - 4 = 0\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
(iv) \(2x^2 + x + 4 = 0\)

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Answer :

(i) \(2x^2 - 7x + 3 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2,
\(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
On dividing the middle term of the equation by 2x, we get the expression :
\(({{7}\over{2}}x)({{1}\over{2x}}) = {{7}\over{4}}\)
On adding and subtracting square of \({{7}\over{4}}\) from the equation \(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
\(x^2 - {{7}\over{2}}x + {{3}\over{2}} + ({{7}\over{4}})^2 - ({{7}\over{4}})^2 = 0\)
\(x^2 + ({{7}\over{4}})^2 - {{7}\over{2}}x + {{3}\over{2}} - ({{7}\over{4}})^2 = 0\)
Using identity {\((a-b)^2 = a^2 + b^2 -2ab\)}
\((x - {{7}\over{4}})^2 + {{24 - 49}\over{16}} = 0\)
\((x - {{7}\over{4}})^2 = {{49 - 24}\over{16}}\)
\((x - {{7}\over{4}})^2 = {{25}\over{16}}\)
\((x - {{7}\over{4}})^2 = ({{\pm5}\over{4}})^2\)
On taking square root on both sides:
\(x - {{7}\over{4}} = {{5}\over{4}}\) and \({{-5}\over{4}}\)
\(x = {{5}\over{4}} + {{7}\over{4}} = {{12}\over{4}} = 3\) and \({{-5}\over{4}} - {{7}\over{4}} = {{2}\over{4}} = {{1}\over{2}}\)
Hence, \(x = {{1} \over {2}} , 3\)

(ii) \(2x^2 + x - 4 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2
\(x^2 + {{x}\over{2}} - 2 = 0\)
On dividing the middle term of the equation by 2x, we get the expression :
\(({{x}\over{2}})({{1}\over{2x}}) = {{1}\over{4}}\)
On adding and subtracting square of \({{1}\over{4}}\) from the equation \(x^2 + {{x}\over{2}} - 2 = 0\)
\(x^2 + {{x}\over{2}} - 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)
\(x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} - 2 - ({{1}\over{16}}) = 0\)
Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}
\((x + {{1}\over{4}})^2 - {{33}\over{16}} = 0\)
\((x + {{1}\over{4}})^2 = {{33}\over{16}}\)
On taking square root both sides:
\(x + {{1}\over{4}} = {{\sqrt{33}}\over{4}}\) and \({{-\sqrt{33}}\over{4}}\)
\(x = {{\sqrt{33}}\over{4}} - {{1}\over{4}} = {{\sqrt{33} - 1}\over{4}}\) and \({{-\sqrt{33}}\over{4}} - {{1}\over{4}} = {{2}\over{4}} = {{-\sqrt{33} - 1}\over{4}}\)
Hence, \(x = {{\sqrt{33} - 1}\over{4}} , {{-\sqrt{33} - 1}\over{4}}\)

(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 4
\(x^2 + \sqrt{3}x + {{3} \over {4}} = 0\)
On dividing the middle term of the equation by 2x, we get the expression :
\( (\sqrt{3}x)({{1}\over{2x}}) \) = \({{\sqrt{3}}\over{2}}\)
On adding and subtracting square of \({{\sqrt{3}}\over{2}}\) (half of the middle term) from the equation \(x^2 + \sqrt{3}x + {{3} \over {4}} = 0\)
\(x^2 + \sqrt{3}x + {{3} \over {4}} + ({{\sqrt{3}}\over{2}})^2 - ({{\sqrt{3}}\over{2}})^2 = 0\)
\(x^2 + ({{3}\over{4}})^2 - \sqrt{3}x + {{3} \over {4}} - ({{3}\over{4}}) = 0\)
Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}
\((x + {{\sqrt{3}}\over{2}})^2 = 0\)
\((x + {{\sqrt{3}}\over{2}}) (x + {{\sqrt{3}}\over{2}}) = 0\)
On taking square root both sides:
\(x + {{\sqrt{3}}\over{2}} = 0\) and \(x + {{\sqrt{3}}\over{2}} = 0\)
\(x = -{{\sqrt{3}}\over{2}}\) and \(x = -{{\sqrt{3}}\over{2}}\)

(iv) \(2x^2 + x + 4 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2
\(x^2 + {{x}\over{2}} + 2 = 0\)
On dividing the middle term of the equation by 2x, we get the expression :
\(({{x}\over{2}})({{1}\over{2x}}) = {{1}\over{4}}\)
On adding and subtracting square of \({{1}\over{4}}\) from the equation \(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
\(x^2 + {{x}\over{2}}x + 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)
\(x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} + 2 - ({{1}\over{4}})^2 = 0\)
Using identity {\((a + b)^2 = a^2 + b^2 -2ab\)}
\((x + {{1}\over{4}})^2 + 2 - {{1}\over{16}} = 0\)
\((x + {{1}\over{4}})^2 = {{1}\over{16}} - 2 = {{1 - 32}\over{16}}\)
\((x + {{1}\over{4}})^2 = \frac{-31}{16}\)
On taking square root on both sides,
Right hand side does not exist since square root of negative number does not exist.
Therefore, there is no solution.

Q2. Find the roots of the following Quadratic Equations by applying quadratic formula.
(i) \(2x^2 - 7x + 3 = 0\)
(ii) \(2x^2 + x - 4 = 0\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
(iv) \(2x^2 + x + 4 = 0\)

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Answer :

(i) \(2x^2 - 7x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{7 ± \sqrt{(-7)^2 - 4(2)(3)}} \over {(2)(2)}}\)
=>\(x = {{7 ± \sqrt{49 - 24}} \over {4}}\)
=>\(x = {{7 ± 5} \over {4}}\)
=>\(x = {{7 + 5} \over {4}} , {{7 - 5} \over {4}}\)
=>\(x = 3 , {{1} \over {2}}\)

(ii) \(2x^2 + x - 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{33}} \over {4}}\)
=>\(x = {{-1 + \sqrt{33}} \over {4}} , {{-1 - \sqrt{33}} \over {4}}\)

(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{(4\sqrt{3})^2 - 4(4)(3)}} \over {(2)(4)}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{0}} \over {8}}\)
=>\(x = {{-\sqrt{3}} \over {2}} , {{-\sqrt{3}} \over {2}}\)

(iv) \(2x^2 + x + 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{-31}} \over {4}}\)
since square root of negative number does not exist.
Therefore, there is no solution.

Q3. Find the roots of the following equations:
(i) \({{x} - {{1}\over {x}}} = 3, x \neq 0\)
(ii) \({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x \neq -4,7\)

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Answer :

(i)\({{x} - {{1}\over {x}}} = 3 \) where x is not equal to 0
=>\({{x^2 - 1} \over {x}} = 3\)
=>\(x^2 - 1 = 3x\)
=>\(x^2 -3x - 1 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(-1)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{13}} \over {2}}\)
=>\(x = {{3 + \sqrt{13}} \over {2}} , {{3 - \sqrt{13}} \over {2}}\)

(ii)\({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x ? -4,7\)
=>\({{(x - 7) - (x + 4)} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\({{-11} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\(-30 = x^2 - 7x + 4x -28\)
=>\(x^2 -3x + 2 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(2)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{1}} \over {2}}\)
=>\(x = {{3 + \sqrt{1}} \over {2}} , {{3 - \sqrt{1}} \over {2}}\)
=>\(x = 2,1\)

Q4. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 13. Find his present age.

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Answer :

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:
\({{1} \over {x - 3}} + {{1} \over {x + 5}} = {{1} \over {3}}\)
=>\({{(x + 5) + (x - 3)} \over {(x - 3)(x + 5)}} = {{1} \over {3}}\)
=>\( 3 (2x + 2) = (x - 3) (x + 5)\)
=>\( 6x + 6 = x^2 - 3x + 5x -15\)
=>\(x^2 - 4x - 15 - 6 = 0\)
=>\(x^2 - 4x -21 = 0\)
Comparing quadratic equation \(x^2 - 4x -21 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?4 and c = ?21
Using quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{4 ± \sqrt{(4)^2 - 4(1)(-21)}} \over {(2)(1)}}\)
=>\(x = {{4 ± \sqrt{16 + 84}} \over {2}}\)
=>\(x = {{4 + 10} \over {2}} , {{4 - 10} \over {2}}\)
=>\(x = 7, -3\)
We discard x=-3. Since age cannot be in negative.
Therefore, present age of Rehman is 7 years.

Q5. 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.

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Answer :

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:
\((x + 2) (27 - x) = 210\)
=>\(27x - x^2 + 54 - 2x = 210\)
=>\(x^2 - 25x + 156 = 0\)
Comparing quadratic equation \(x^2 - 25x + 156 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?25 and c = 156
Applying Quadratic Formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{25 ± \sqrt{(25)^2 - 4(1)(156)}} \over {(2)(1)}}\)
=>\(x = {{25 ± \sqrt{625 - 624}} \over {2}}\)
=>\(x = {{25 + 1} \over {2}} , {{25 - 1} \over {2}}\)
=>\(x = 13, 12\)
Therefore, Shefali’s marks in Mathematics = 13 or 12
Shefali’s marks in English = 30 – x = 30 – 13 = 17
Or Shefali’s marks in English = 30 – x = 30 – 12 = 18
Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).

Q6. 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.

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Answer :

Let shorter side of rectangle = x metres
Let diagonal of rectangle = (x + 60) metres
Let longer side of rectangle = (x + 30) metres
According to pythagoras theorem,
\((x + 60)^2 = (x + 30)^2 + x^2\)
=>\( x^2 + 3600 + 120x = x^2 + 900 + 60x + x^2\)
=>\(x^2 - 60x - 2700 = 0\)
Comparing equation with \(x^2 - 60x - 2700 = 0\) standard form \(ax^2 + bx + c = 0\),
We get a = 1, b = - 60 and c = - 2700
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{60 ± \sqrt{(60)^2 - 4(1)(-2700)}} \over {(2)(1)}}\)
=>\(x = {{60 ± \sqrt{3600 + 10800}} \over {2}} = {{60 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{60 + 120} \over {2}} , {{60 - 120} \over {2}}\)
=>\(x = 90,-30\)
We ignore –30. Since length cannot be in negative.
Therefore, x = 90 which means length of shorter side = 90 metres
And length of longer side = x + 30 = 90 + 30 = 120 metres
Therefore, length of sides are 90 and 120 in metres.

Q7. 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.

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Answer :

Let smaller number = x and let larger number = y
According to condition:
=>\(y^2 - x^2 = 180\)… (1)
Also, we are given that square of smaller number is 8 times the larger number.
=>\(x^2 = 8y\) … (2)
Putting equation (2) in (1), we get
\(y^2 - 8y = 180\)
=>\(y^2 - 8y - 180 = 0\)
Comparing equation \(y^2 - 8y - 180 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?8 and c = ?180
Using quadratic formula = \(y = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(y = {{8 ± \sqrt{(-8)^2 - 4(1)(-180)}} \over {(2)(1)}}\)
=>\(y = {{8 ± \sqrt{64 + 720}} \over {2}} = {{60 ± \sqrt{784}} \over {2}}\)
=>\(y = {{8 + 28} \over {2}} , {{8 - 28} \over {2}}\)
=>\(y = 18,-10\)
Using equation (2) to find smaller number:
\(x^2 = 8y\)
=>\(x^2 = 8y = (8)(18) = 144\)
=>\( x = ±12\)
And, = \(x^2 = 8y = (8)(-10) = -80\) {No real solution for x}
Therefore, two numbers are (12, 18) or (-12, 18)

Q8. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

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Answer :

Let the speed of the train = x km/hr
If, speed had been 5 km/hr more, train would have taken 1 hour less.
So, according to this condition
\({{360} \over {x}} = {{360} \over {x + 5}} + 1\)
=>\(360 ( {{1}\over{x}} - {{1}\over {x + 5}}) = 1\)
=> \(360 ( {{x + 5 - x}\over{x(x + 5)}}) = 1\)
=>\((360)(5) = x^2 + 5x\)
=>\(x^2 + 5x - 1800 = 0\)
Comparing equation \(x^2 + 5x - 1800 = 0\) with general equation \(ax^2 + bx + c = 0\),
We get a = 1, b = 5 and c = -1800
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-5 ± \sqrt{(5)^2 - 4(1)(-1800)}} \over {(2)(1)}}\)
=>\(x = {{-5 ± \sqrt{25 + 7200}} \over {2}} = {{-5 ± \sqrt{7225}} \over {2}}\)
=>\(x = {{-5 + 85} \over {2}} , {{-5 - 85} \over {2}}\)
=>\(x = 40,-45\)
Since speed of train cannot be in negative. Therefore, we discard x = -45
Therefore, speed of train = 40 km/hr

Q9. Two water taps together can fill a tank \({9}{\dfrac{3}{8}}\)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.

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Answer :

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 \({{1} \over {x}} th\) part of tank in 1 hour.… (1)
And, tap of larger diameter fills \({{1} \over {x - 10}} th\) part of tank in 1 hour. … (2)
When two taps are used together, they fill tank in 758 hours.
In 1 hour \({{8} \over {75}} th\), they fill part of tank \({{1} \over {{{75} \over {8}}}} = {{8} \over {75}}\)… (3)
From (1), (2) and (3),
=>\( {{1}\over{x}} + {{1}\over {x - 10}} = {{8} \over {75}}\)
=> \( {{x - 10 + x}\over{x(x - 10)}} = {{8} \over {75}}\)
=>\( 75 (2x - 10) = 8 (x^2 - 10x)\)
=> \(150x – 750 = 8x^2 - 80x\)
=>\(8x^2 - 80x - 150x +750 = 0\)
=>\(4x^2 - 115x + 375 = 0\)
Comparing equation \(4x^2 - 115x + 375 = 0\) with general equation \(ax^2 + bx + c = 0\),
We get a = 4, b = -115 and c = 375
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{115 ± \sqrt{(-115)^2 - 4(4)(375)}} \over {(2)(4)}}\)
=>\(x = {{115 ± \sqrt{13225 - 6000}} \over {8}} = {{115 ± \sqrt{7225}} \over {8}}\)
=>\(x = {{115 + 85} \over {8}} , {{115 - 85} \over {8}}\)
=>\(x = 25,3.75\)
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.

Q10. 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 11 km/h more than that of the passenger train, find the average speed of two trains.

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Answer :

Let average speed of passenger train = x km/h
Let average speed of express train = (x + 11) km/h
Time taken by passenger train to cover 132 km = \({{132} \over {x}}\) hours
Time taken by express train to cover 132 km = \({{132} \over {x + 11}}\) hours
According to the given condition,
\({{132} \over {x}} = {{132} \over {x + 11}} + 1\)
=>\(132({{1} \over {x}} - {{1} \over {x + 11}}) = 1\)
=>\(132({{x + 11 - x} \over {x(x + 11)}}) = 1\)
=> \(132 (11) = x (x + 11)\)
=>\(1452 = x^2 + 11x\)
=>\(x^2 + 11x -1452 = 0\)
Comparing equation \(x^2 + 11x -1452 = 0\) with general quadratic equation \(ax^2 + bx + c = 0\), we get a = 1, b = 11 and c = ?1452
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-11 ± \sqrt{(11)^2 - 4(1)(-1452)}} \over {(2)(1)}}\)
=>\(x = {{-11 ± \sqrt{121 + 5808}} \over {2}} = {{-11 ± \sqrt{5929}} \over {2}}\)
=>\(x = {{-11 + 77} \over {2}} , {{-11 - 77} \over {2}}\)
=>\(x = 33, -44\)
As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h
And, speed of express train = x + 11 = 33 + 11 = 44 km/h

Q11. Sum of areas of two squares is 468 \(m^2\). If, the difference of their perimeters is 24 metres, find the sides of the two squares.

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Answer :

Let perimeter of first square = x metres
Let perimeter of second square = (x + 24) metres
Length of side of first square = \({{x} \over {4}}\) metres {Perimeter of square = 4 × length of side}
Length of side of second square = \({{x + 24} \over {4}}\) metres
Area of first square = side × side = \({{x} \over {4}} × {{x} \over {4}} = {{x^2} \over {16}} m^2\)
Area of second square = \(({{x + 24} \over {4}})^2\)
According to given condition:
\({{x^2} \over {16}} + ({{x + 24} \over {4}})^2 = 468\)
=>\({{x^2} \over {16}} + {{x^2 + 576 + 48x} \over {16}} = 468\)
=>\({{x^2 + x^2 + 576 + 48x} \over {16}} = 468\)
=> \(2x^2 + 576 + 48x = (468)(16)\)
=> \(2x^2 + 576 + 48x = 7488\)
=>\(2x^2 + 48x - 6912 = 0\)
=>\(x^2 + 24x - 3456 = 0\)
Comparing equation \(x^2 + 24x - 3456 = 0\) with standard form \(ax^2 + bx + c = 0\),
We get a = 1, b =24 and c = -3456
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-24 ± \sqrt{(24)^2 - 4(1)(-3456)}} \over {(2)(1)}}\)
=>\(x = {{-24 ± \sqrt{576 + 13824}} \over {2}} = {{-24 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{-24 + 120} \over {2}} , {{-24 - 120} \over {2}}\)
=>\(x = 48, -72\)
Perimeter of square cannot be in negative. Therefore, we discard x=-72.
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres
=> Side of First square = \({{Perimeter} \over {4}} = {{48} \over {4}} = 12\) m
And, Side of second Square = \({{Perimeter} \over {4}} = {{72} \over {4}} = 18\) m

Solution for Exercise 4.4

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.
(i)\(2x^2 - 3x + 5 = 0\)
(ii)\(3x^2 -4 \sqrt{3} x + 4 = 0\)
(iii)\(2x^2 - 6x + 3 = 0\)

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Answer :

Ans.(i)\(2x^2 - 3x + 5 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 2, b = ?3 and c = 5
Discriminant = \(b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 – 40 = ?31\)
Discriminant is less than 0 which means equation has no real roots.

(ii)\(3x^2 -4 \sqrt{3} x + 4 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 3, b = \(-4 \sqrt{3}\) and c = 4 Discriminant = \(b^2 - 4ac = (-4 \sqrt{3})^2 - 4(3)(4) = 48 – 48 = 0\)
Discriminant is equal to zero which means equations has equal real roots.
Applying quadratic \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to find roots,
=>\(x = {{4 \sqrt{3} ± \sqrt{0}} \over {6}} = {{2 \sqrt{3}} \over {3}} \)
Because, equation has two equal roots, it means x = \({{2 \sqrt{3}} \over {3}} ,{{2 \sqrt{3}} \over {3}} \)

(iii)\(2x^2 - 6x + 3 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 2, b = ?6, and c = 3
Discriminant =\(b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 – 24 = 12\)
Value of discriminant is greater than zero.
Therefore, equation has distinct and real roots.
Applying quadratic \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to find roots,
\(x = {{6 ± \sqrt{12}} \over {4}} = {{6 ± 2 \sqrt{3}} \over {4}} = {{3 ± \sqrt{3}} \over {2}}\) \(x = {{3 + \sqrt{3}} \over {2}} , {{3 - \sqrt{3}} \over {2}}\)

Q2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.
(i) \(2x^2 + kx + 3 = 0\)
(ii) \(kx (x - 2) + 6 = 0\)

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Answer :

Ans.(i)\(2x^2 + kx + 3 = 0\)
We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.
Comparing equation with \(2x^2 + kx + 3 = 0\) general quadratic equation \(ax^2 + bx + c = 0\), we get a = 2, b = k and c = 3
Discriminant \( b^2 - 4ac = k^2 - 4 (2) (3) = k^2 - 24\)
Putting discriminant equal to zero
\(k^2 - 24= 0\)
\( k^2 = 24 \)
\( k = ± \sqrt{2} = ± 2 \sqrt{6}\)

(ii)\(kx (x - 2) + 6 = 0\)
\( kx^2 - 2kx + 6 = 0\)
Comparing quadratic equation \(kx (x - 2) + 6 = 0\) with general form \(ax^2 + bx + c = 0\), we get a = k, b = ?2k and c = 6
Discriminant \( b^2 - 4ac = (-2k)^2 - 4 (k) (6) = 4k^2 - 24k\)
We know that two roots of quadratic equation are equal only if discriminant is equal to zero.
Putting discriminant equal to zero
\(4k^2 - 24k = 0\)
\(4k (k - 6) = 0=> k = 0, 6\)
The basic definition of quadratic equation says that quadratic equation is the equation of the form \(ax^2 + bx + c = 0\), where a ? 0.
Therefore, in equation , we cannot have k = 0.
Therefore, we discard k = 0.
Hence the answer is k = 6.

Q3. 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.

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Answer :

Ans.Let breadth of rectangular mango grove = x metres
Let length of rectangular mango grove = 2x metres
Area of rectangle = length × breadth = \(x × 2x = 2x^2 m^2\)
According to given condition:
\(2x^2 = 800\)
\(2x^2 - 800 = 0 => x^2 - 400 = 0\)
Comparing equation \( x^2 - 400 = 0\) with general form of quadratic equation \(ax^2 + bx + c = 0\) ,
we get a = 1, b = 0 and c = -400
Discriminant = \( b^2 -4ac = (0)^2 - 4 (1) (-400) = 1600\)
Discriminant is greater than 0 means that equation has two distinct real roots.
Therefore, it is possible to design a rectangular grove.
Applying quadratic formula,\(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to solve equation,
\(x = {{0 ± \sqrt{1600}} \over {2(1)}}\) = ± {{40} \over {2} } = ± 20\)
x = 20, -20
We discard negative value of x because breadth of rectangle cannot be in negative.
Therefore, x = breadth of rectangle = 20 metres
Length of rectangle = 2x = 40 metres

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.

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Answer :

Ans. Let age of first friend = x years and let age of second friend = (20 - x) years
Four years ago, age of first friend = (x - 4) years
Four years ago, age of second friend = (20 - x) - 4 = (16 - x) years
According to given condition,
\((x - 4) (16 - x) = 48\)
=>\(16x - x^2 - 64 + 4x = 48\)
=>\(20x - x^2 - 112 = 0\)
=>\(x^2 - 20x + 112 = 0\)
Comparing equation , \(x^2 - 20x + 112 = 0\) with general quadratic equation \(ax^2 + bx + c = 0\),
we get a = 1, b = -20 and c = 112
Discriminant = \( b^2 - 4ac = (-20)^2 - 4 (1) (112) = 400 – 448 = -48 < 0\)
Discriminant is less than zero which means we have no real roots for this equation.
Therefore, the give situation is not possible.

Q5. Is it possible to design a rectangular park of perimeter 80 metres and area 400 \(m^2\). If so, find its length and breadth.

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Answer :

Ans. Let length of park = x metres
We are given area of rectangular park = 400 \(m^2\)
Therefore, breadth of park = \({{400} \over {x}}\) metres{Area of rectangle = length × breadth}
Perimeter of rectangular park = 2 (length + breath) = \( 2 ( x + {{400} \over {x}})\) metres
We are given perimeter of rectangle = 80 metres
According to condition:
\( 2 ( x + {{400} \over {x}}) = 80 \)
\( 2 ( {{x^2 + 400} \over {x}})\)
\(2x^2 + 800 = 80x\)
\(2x^2 - 80x + 800 =0\)
\( x^2 - 40x + 400 = 0\)
Comparing equation, \( x^2 - 40x + 400 = 0\) with general quadratic equation \( ax^2 + bx + c = 0\),
we get a = 1, b = -40 and c = 400
Discriminant = \( b^2 - 4ac = (-40)^2 - 4 (1) (400) = 1600 – 1600 = 0\)
Discriminant is equal to 0.
Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area 400 \(m^2\).
Using quadratic formula \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to solve equation,
\(x = {{40 ± \sqrt{0}} \over {2}} = {{40} \over {2}} = 20\)
Here, both the roots are equal to 20.
Therefore, length of rectangular park = 20 metres
Breadth of rectangular park = \({{400} \over {x}} = {{400} \over {20}} = 20 m \)