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)\(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

(i)\((x + 1)^2 = 2(x - 3)\)

Using identity \((a + b)^2 = a^2 + 2ab + b^2\)

\( \Rightarrow x^2 + 1 + 2x = 2x - 6\)

\( \Rightarrow x^2 + 7 = 0\)

Hence it is a quadratic equation since the degree is 2.

(ii)\(x^2 - 2x = (-2) (3 - x)\)

\( \Rightarrow x^2 - 2x = -6 + 2x\)

\( \Rightarrow x^2 - 2x - 2x + 6 = 0\)

\( \Rightarrow x^2 - 4x + 6 = 0\)

Hence it is a quadratic equation with degree 2.

(iii)\((x - 2) (x + 1) = (x - 1) (x + 3)\)

\( \Rightarrow x^2 + x - 2x - 2 = x^2 + 3x - x - 3\)

\( \Rightarrow x^2 + x - 2x - 2 - (x^2 + 3x - x - 3) = 0\)

\( \Rightarrow x^2 - x^2 + x - 2x - 3x + x - 2 + 3 = 0\)

\( \Rightarrow -3x + 1 = 0\)

Hence it is a not quadratic equation since the degree is 1.

(iv)\((x - 3) (2x + 1) = x (x + 5)\)

\( \Rightarrow 2x^2 + x - 6x - 3 = x^2 + 5x\)

\( \Rightarrow 2x^2 + x - 6x - 3 - x^2 - 5x = 0\)

\( \Rightarrow x^2 - 10x - 3 = 0\)

Hence it is a quadratic equation since the degree is 2.

(v)\((2x - 1) (x - 3) = (x + 5) (x - 1)\)

\( \Rightarrow 2x^2 - x - 6x + 3 = x^2 - x + 5x - 5\)

\( \Rightarrow 2x^2 - 7x + 3 - x^2 + x - 5x + 5 = 0\)

\( \Rightarrow x^2 - 10x - 3 = 0\)

Hence it is a quadratic equation since the degree is 2.

(vi)\(x^2 + 3x + 1 = (x - 2)^2\)

Using identity \((a - b)^2 = a^2 - 2ab + b^2\)

\( \Rightarrow x^2 + 3x + 1 = x^2 - 4x + 4 \)

\( \Rightarrow x^2 + 3x + 1 - x^2 + 4x - 4 = 0\)

\( \Rightarrow 7x - 3 = 0\)

Hence it is a not quadratic equation since the degree is 1.

(vii)\((x + 2)^3 = 2x(x^2 - 1)\)

Using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)

\( \Rightarrow x^3 + 2^3 + 3(x)(2)(x + 2) = 2x(x^2 - 1)\)

\( \Rightarrow x^3 + 8 + 6x(x + 2) = 2x^3 - 2x\)

\( \Rightarrow 2x^3 - 2x - x^3 - 8 - 6x^2 - 12x = 0\)

\( \Rightarrow x^3 - 6x^2 - 14x - 8 = 0\)

Hence it is a not quadratic equation since the degree is 3

(viii)\(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

Using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)

\( \Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3(x)(2)(x-2)\)

\( \Rightarrow -4x^2 - x + 1 = -8 - 6x^2 + 12x\)

\( \Rightarrow 2x^2 - 13x + 9 = 0\)

Hence it is a quadratic equation since the degree is 2

Q2 )
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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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.

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) \(100x^2 - 20x + 1 = 0\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

** (i) ** \(x^2 -3x - 10 = 0\)

(Splitting * -3x * as * 2x - 5x*)

\(\Rightarrow x^2 + 2x - 5x - 10 = 0 \)

\( \Rightarrow x(x + 2) - 5(x + 2) = 0 \)

\( \Rightarrow (x - 5)(x + 2) = 0 \)

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

\( \Rightarrow 2x^2 +4x - 3x - 6 = 0\)

\( \Rightarrow 2x(x + 2) - 3(x + 2) = 0 \)

\( \Rightarrow (2x - 3)(x + 2) = 0 \)

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,

(iii)

(Splitting

\( \Rightarrow \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \)

\( \Rightarrow \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \)

\( \Rightarrow (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \)

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,

(iv)

Multiplying and dividing the entire equation with 8, we get -

\(\frac{1}{8}(16x^2 - 8x + 1) = 0\)

(Splitting

\( \Rightarrow \frac{1}{8}(16x^2 - 4x - 4x + 1) = 0\)

\( \Rightarrow \frac{1}{8}[4x(4x - 1) - 1(4x - 1)] = 0 \)

\( \Rightarrow \frac{1}{8}(4x - 1)(4x - 1) = 0 \)

\( \Rightarrow \frac{1}{8}(4x - 1)^2= 0 \)

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,

(v)

(Splitting

\(\Rightarrow 100 x^2 – 10x - 10x + 1 = 0\)

\( \Rightarrow 10x(10x - 1) - 1(10x - 1) = 0 \)

\( \Rightarrow (10x - 1)(10x - 1) = 0 \)

\( \Rightarrow (10x - 1)^2 = 0 \)

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 Rs.750. We would like to find out the number of toys produced on that day.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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,

\(\Rightarrow (x - 5)(40 - x) = 124\)

\(\Rightarrow 40x - x^2 - 200 + 5x = 124\)

\( \Rightarrow -x^2 + 45x - 324 = 0\)

\(\Rightarrow x^2 - 45x + 324 = 0\)

(Splitting * -45x * as * -9x - 36x *)

\(\Rightarrow x^2 - 9x - 36x + 324 = 0\)

\(\Rightarrow x(x - 9) - 36(x - 9) = 0\)

\(\Rightarrow (x - 36)(x - 9) = 0\)

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,

\(\Rightarrow x(55 - x) = 750\)

\(\Rightarrow 55x - x^2 - 750 = 0\)

\(\Rightarrow x^2 - 55x + 750 = 0\)

(Splitting

\(\Rightarrow x^2 - 30x - 25x + 750 = 0\)

\(\Rightarrow x(x - 30) - 25(x - 30) = 0\)

\(\Rightarrow (x - 25)(x - 30) = 0\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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,

\(\Rightarrow x(27 - x) = 182\)

\(\Rightarrow 27x - x^2 - 182 = 0\)

\(\Rightarrow x^2 - 27x + 182 = 0\)

(Splitting * -27x * as * -13x - 14x *)

\(\Rightarrow x^2 - 13x - 14x + 182 = 0\)

\(\Rightarrow x(x - 13) - 14(x - 13) = 0\)

\(\Rightarrow (x - 14)(x - 13) = 0\)

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. **

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

Let the first number be \(x\)

The second number can be represented by \(x + 1\).

From the given condition,

\(\Rightarrow x^2 + (x + 1)^2 = 365\)

\(\Rightarrow x^2 + x^2 + 2x + 1 - 365 = 0\)

\(\Rightarrow 2x^2 + 2x -364 = 0\)

\(\Rightarrow x^2 + x - 182 = 0\)

(Splitting * x * as * 14x - 13x *)

\(\Rightarrow x^2 + 14x - 13x - 182 = 0\)

\(\Rightarrow x(x + 14) - 13(x + 14) = 0\)

\(\Rightarrow (x - 13)(x + 14) = 0\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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)

\(\Rightarrow 13^2 = x^2 + (x - 7)^2 \)

\(\Rightarrow 169 = x^2 + x^2 - 14x + 49 \)

\(\Rightarrow 2x^2 - 14x - 120 = 0 \)

\(\Rightarrow x^2 - 7x - 60 = 0\)

(Splitting * -7x * as * 5x - 12x *)

\(\Rightarrow x^2 + 5x - 12x - 60 = 0\)

\(\Rightarrow x(x + 5) - 12(x + 5) = 0\)

\(\Rightarrow (x - 12)(x + 5) = 0\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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.,

\(\Rightarrow x(2x + 3) = 90\)

\(\Rightarrow 2x^2 + 3x - 90 = 0\)

(Splitting * 3x * as * -12x + 15x *)

\(\Rightarrow 2x^2 + -12x +15x - 90 = 0\)

\(\Rightarrow 2x(x - 6) + 15(x - 6) = 0\)

\(\Rightarrow(2x + 15)(x - 6) = 0\)

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.
**

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) \(2x^2 + x + 4 = 0\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

\(\Rightarrow x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)

\(\Rightarrow x^2 - {{7}\over{2}}x + {{3}\over{2}} + ({{7}\over{4}})^2 - ({{7}\over{4}})^2 = 0\)

\(\Rightarrow x^2 + ({{7}\over{4}})^2 - {{7}\over{2}}x + {{3}\over{2}} - ({{7}\over{4}})^2 = 0\)

Using identity {\((a-b)^2 = a^2 + b^2 -2ab\)}

\(\Rightarrow (x - {{7}\over{4}})^2 + {{24 - 49}\over{16}} = 0\)

\(\Rightarrow (x - {{7}\over{4}})^2 = {{49 - 24}\over{16}}\)

\(\Rightarrow (x - {{7}\over{4}})^2 = {{25}\over{16}}\)

\(\Rightarrow (x - {{7}\over{4}})^2 = ({{\pm5}\over{4}})^2\)

On taking square root on both sides:

\( \Rightarrow x - {{7}\over{4}} = {{5}\over{4}}\) and \({{-5}\over{4}}\)

\(\Rightarrow x = {{5}\over{4}} + {{7}\over{4}} = {{12}\over{4}} = 3\) and

\( \Rightarrow x= {{-5}\over{4}} + {{7}\over{4}} = {{2}\over{4}} = {{1}\over{2}}\)

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 \(x^2 + {{x}\over{2}} - 2 = 0\)

\(\Rightarrow x^2 + {{x}\over{2}} - 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)

\(\Rightarrow x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} - 2 - ({{1}\over{16}}) = 0\)

Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}

\(\Rightarrow (x + {{1}\over{4}})^2 - {{33}\over{16}} = 0\)

\(\Rightarrow (x + {{1}\over{4}})^2 = {{33}\over{16}}\)

On taking square root both sides:

\(\Rightarrow x + {{1}\over{4}} = {{\sqrt{33}}\over{4}}\) and \({{-\sqrt{33}}\over{4}}\)

\(\Rightarrow x = {{\sqrt{33}}\over{4}} - {{1}\over{4}} = {{\sqrt{33} - 1}\over{4}}\) 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

\(\Rightarrow x^2 + \sqrt{3}x + {{3} \over {4}} = 0\)

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 \(x^2 + \sqrt{3}x + {{3} \over {4}} = 0\)

\(\Rightarrow x^2 + \sqrt{3}x + {{3} \over {4}} + ({{\sqrt{3}}\over{2}})^2 - ({{\sqrt{3}}\over{2}})^2 = 0\)

\(\Rightarrow x^2 + ({{3}\over{4}})^2 - \sqrt{3}x + {{3} \over {4}} - ({{3}\over{4}}) = 0\)

Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}

\(\Rightarrow (x + {{\sqrt{3}}\over{2}})^2 = 0\)

\(\Rightarrow (x + {{\sqrt{3}}\over{2}}) (x + {{\sqrt{3}}\over{2}}) = 0\)

On taking square root both sides:

\(\Rightarrow x + {{\sqrt{3}}\over{2}} = 0\) and \(x + {{\sqrt{3}}\over{2}} = 0\)

\(\Rightarrow x = -{{\sqrt{3}}\over{2}}\) 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

\(\Rightarrow x^2 + {{x}\over{2}} + 2 = 0\)

On dividing the middle term of the equation by 2x, we get the expression :

\(\Rightarrow({{x}\over{2}})({{1}\over{2x}}) = {{1}\over{4}}\)

On adding and subtracting square of \({{1}\over{4}}\) from the equation \(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)

\(\Rightarrow x^2 + {{x}\over{2}}x + 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)

\(\Rightarrow x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} + 2 - ({{1}\over{4}})^2 = 0\)

Using identity {\((a + b)^2 = a^2 + b^2 -2ab\)}

\(\Rightarrow (x + {{1}\over{4}})^2 + 2 - {{1}\over{16}} = 0\)

\(\Rightarrow (x + {{1}\over{4}})^2 = {{1}\over{16}} - 2 = {{1 - 32}\over{16}}\)

\(\Rightarrow (x + {{1}\over{4}})^2 = \frac{-31}{16}\)

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) \(2x^2 + x + 4 = 0\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

(i) \(2x^2 - 7x + 3 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \) \(x = {{7 ± \sqrt{(-7)^2 - 4(2)(3)}} \over {(2)(2)}}\)

\(\Rightarrow \)\(x = {{7 ± \sqrt{49 - 24}} \over {4}}\)

\(\Rightarrow \)\(x = {{7 ± 5} \over {4}}\)

\(\Rightarrow \)\(x = {{7 + 5} \over {4}} , {{7 - 5} \over {4}}\)

\(\Rightarrow \)\(x = 3 , {{1} \over {2}}\)

(ii) \(2x^2 + x - 4 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{33}} \over {4}}\)

\(\Rightarrow \)\(x = {{-1 + \sqrt{33}} \over {4}} , {{-1 - \sqrt{33}} \over {4}}\)

(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-4\sqrt{3} ± \sqrt{(4\sqrt{3})^2 - 4(4)(3)}} \over {(2)(4)}}\)

\(\Rightarrow \)\(x = {{-4\sqrt{3} ± \sqrt{0}} \over {8}}\)

\(\Rightarrow \)\(x = {{-\sqrt{3}} \over {2}} , {{-\sqrt{3}} \over {2}}\)

(iv) \(2x^2 + x + 4 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{-31}} \over {4}}\)

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) \({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x \neq -4,7\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

(i)\({{x} - {{1}\over {x}}} = 3 \) where x is not equal to 0

\(\Rightarrow \) \({{x^2 - 1} \over {x}} = 3\)

\(\Rightarrow \)\(x^2 - 1 = 3x\)

\(\Rightarrow \)\(x^2 -3x - 1 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{3 ± \sqrt{(3)^2 - 4(1)(-1)}} \over {(2)(1)}}\)

\(\Rightarrow \)\(x = {{3 ± \sqrt{13}} \over {2}}\)

\(\Rightarrow \)\(x = {{3 + \sqrt{13}} \over {2}} , {{3 - \sqrt{13}} \over {2}}\)

(ii)\({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x ? -4,7\)

\(\Rightarrow \)\({{(x - 7) - (x + 4)} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)

\(\Rightarrow \)\({{-11} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)

\(\Rightarrow \)\(-30 = x^2 - 7x + 4x -28\)

\(\Rightarrow \)\(x^2 -3x + 2 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{3 ± \sqrt{(3)^2 - 4(1)(2)}} \over {(2)(1)}}\)

\(\Rightarrow \)\(x = {{3 ± \sqrt{1}} \over {2}}\)

\(\Rightarrow \)\(x = {{3 + \sqrt{1}} \over {2}} , {{3 - \sqrt{1}} \over {2}}\)

\(\Rightarrow \)\(x = 2,1\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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:

\(\Rightarrow \) \({{1} \over {x - 3}} + {{1} \over {x + 5}} = {{1} \over {3}}\)

\(\Rightarrow \)\({{(x + 5) + (x - 3)} \over {(x - 3)(x + 5)}} = {{1} \over {3}}\)

\(\Rightarrow \)\( 3 (2x + 2) = (x - 3) (x + 5)\)

\(\Rightarrow \)\( 6x + 6 = x^2 - 3x + 5x -15\)

\(\Rightarrow \)\(x^2 - 4x - 15 - 6 = 0\)

\(\Rightarrow \)\(x^2 - 4x -21 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{4 ± \sqrt{(4)^2 - 4(1)(-21)}} \over {(2)(1)}}\)

\(\Rightarrow \)\(x = {{4 ± \sqrt{16 + 84}} \over {2}}\)

\(\Rightarrow \)\(x = {{4 + 10} \over {2}} , {{4 - 10} \over {2}}\)

\(\Rightarrow \)\(x = 7, -3\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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:

\(\Rightarrow \) \((x + 2) (27 - x) = 210\)

\(\Rightarrow \) \(27x - x^2 + 54 - 2x = 210\)

\(\Rightarrow \) \(x^2 - 25x + 156 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \) \(x = {{25 ± \sqrt{(25)^2 - 4(1)(156)}} \over {(2)(1)}}\)

\(\Rightarrow \) \(x = {{25 ± \sqrt{625 - 624}} \over {2}}\)

\(\Rightarrow \) \(x = {{25 + 1} \over {2}} , {{25 - 1} \over {2}}\)

\(\Rightarrow \) \(x = 13, 12\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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,

\((x + 60)^2 = (x + 30)^2 + x^2\)

\(\Rightarrow \) \( x^2 + 3600 + 120x = x^2 + 900 + 60x + x^2\)

\(\Rightarrow \)\(x^2 - 60x - 2700 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{60 ± \sqrt{(60)^2 - 4(1)(-2700)}} \over {(2)(1)}}\)

\(\Rightarrow \)\(x = {{60 ± \sqrt{3600 + 10800}} \over {2}} = {{60 ± \sqrt{14400}} \over {2}}\)

\(\Rightarrow \)\(x = {{60 + 120} \over {2}} , {{60 - 120} \over {2}}\)

\(\Rightarrow \)\(x = 90,-30\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

\(y^2 - 8y = 180\)

\(\Rightarrow \) \(y^2 - 8y - 180 = 0\)

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 = \(y = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \) \(y = {{8 ± \sqrt{(-8)^2 - 4(1)(-180)}} \over {(2)(1)}}\)

\(\Rightarrow \) \(y = {{8 ± \sqrt{64 + 720}} \over {2}} = {{60 ± \sqrt{784}} \over {2}}\)

(\Rightarrow \) \(y = {{8 + 28} \over {2}} , {{8 - 28} \over {2}}\)

(\Rightarrow \) \(y = 18,-10\)

Using equation (2) to find smaller number:

\(x^2 = 8y\)

(\Rightarrow \) \(x^2 = 8y = (8)(18) = 144\)

(\Rightarrow \) \( x = ±12\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

\(\Rightarrow {{360} \over {x}} = {{360} \over {x + 5}} + 1\)

\(\Rightarrow 360 ( {{1}\over{x}} - {{1}\over {x + 5}}) = 1\)

\(\Rightarrow 360 ( {{x + 5 - x}\over{x(x + 5)}}) = 1\)

\(\Rightarrow (360)(5) = x^2 + 5x\)

\(\Rightarrow x^2 + 5x - 1800 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow x = {{-5 ± \sqrt{(5)^2 - 4(1)(-1800)}} \over {(2)(1)}}\)

\(\Rightarrow x = {{-5 ± \sqrt{25 + 7200}} \over {2}} = {{-5 ± \sqrt{7225}} \over {2}}\)

\(\Rightarrow x = {{-5 + 85} \over {2}} , {{-5 - 85} \over {2}}\)

\(\Rightarrow x = 40,-45\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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)

\( \Rightarrow {{1}\over{x}} + {{1}\over {x - 10}} = {{8} \over {75}}\)

\( \Rightarrow {{x - 10 + x}\over{x(x - 10)}} = {{8} \over {75}}\)

\(\Rightarrow 75 (2x - 10) = 8 (x^2 - 10x)\)

\(\Rightarrow 150x – 750 = 8x^2 - 80x\)

\(\Rightarrow 8x^2 - 80x - 150x +750 = 0\)

\(\Rightarrow 4x^2 - 115x + 375 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow x = {{115 ± \sqrt{(-115)^2 - 4(4)(375)}} \over {(2)(4)}}\)

\(\Rightarrow x = {{115 ± \sqrt{13225 - 6000}} \over {8}} = {{115 ± \sqrt{7225}} \over {8}}\)

\(\Rightarrow x = {{115 + 85} \over {8}} , {{115 - 85} \over {8}}\)

\(\Rightarrow x = 25,3.75\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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,

\(\Rightarrow {{132} \over {x}} = {{132} \over {x + 11}} + 1\)

\(\Rightarrow 132({{1} \over {x}} - {{1} \over {x + 11}}) = 1\)

\(\Rightarrow 132({{x + 11 - x} \over {x(x + 11)}}) = 1\)

\(\Rightarrow 132 (11) = x (x + 11)\)

\(\Rightarrow 1452 = x^2 + 11x\)

\(\Rightarrow x^2 + 11x -1452 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow x = {{-11 ± \sqrt{(11)^2 - 4(1)(-1452)}} \over {(2)(1)}}\)

\(\Rightarrow x = {{-11 ± \sqrt{121 + 5808}} \over {2}} = {{-11 ± \sqrt{5929}} \over {2}}\)

\(\Rightarrow x = {{-11 + 77} \over {2}} , {{-11 - 77} \over {2}}\)

\(\Rightarrow x = 33, -44\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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:

\(\Rightarrow {{x^2} \over {16}} + ({{x + 24} \over {4}})^2 = 468\)

\(\Rightarrow{{x^2} \over {16}} + {{x^2 + 576 + 48x} \over {16}} = 468\)

\(\Rightarrow{{x^2 + x^2 + 576 + 48x} \over {16}} = 468\)

\(\Rightarrow 2x^2 + 576 + 48x = (468)(16)\)

\(\Rightarrow 2x^2 + 576 + 48x = 7488\)

\(\Rightarrow 2x^2 + 48x - 6912 = 0\)

\(\Rightarrow x^2 + 24x - 3456 = 0\)

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 = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow x = {{-24 ± \sqrt{(24)^2 - 4(1)(-3456)}} \over {(2)(1)}}\)

\(\Rightarrow x = {{-24 ± \sqrt{576 + 13824}} \over {2}} = {{-24 ± \sqrt{14400}} \over {2}}\)

\(\Rightarrow x = {{-24 + 120} \over {2}} , {{-24 - 120} \over {2}}\)

\(\Rightarrow x = 48, -72\)

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

\(\Rightarrow \) 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

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)\(2x^2 - 6x + 3 = 0\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Answer :

(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,

\(\Rightarrow \) \(x = {{4 \sqrt{3} ± \sqrt{0}} \over {6}} = {{2 \sqrt{3}} \over {3}} \)

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) \(kx (x - 2) + 6 = 0\)

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

\(\Rightarrow k^2 - 24= 0\)

\( \Rightarrow k^2 = 24 \)

\(\Rightarrow k = ± \sqrt{2} = ± 2 \sqrt{6}\)

(ii)\(kx (x - 2) + 6 = 0\)

\(\Rightarrow kx^2 - 2kx + 6 = 0\)

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

\(\Rightarrow 4k^2 - 24k = 0\)

\(\Rightarrow 4k (k - 6) = 0\)

\(\Rightarrow k = 0, 6\)

The basic definition of quadratic equation says that quadratic equation is the equation of the form \(ax^2 + bx + c = 0\), where a \(\neq \) 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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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:

\(\Rightarrow 2x^2 = 800\)

\(\Rightarrow 2x^2 - 800 = 0 \)

\(\Rightarrow x^2 - 400 = 0\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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,

\(\Rightarrow(x - 4) (16 - x) = 48\)

\(\Rightarrow 16x - x^2 - 64 + 4x = 48\)

\(\Rightarrow 20x - x^2 - 112 = 0\)

\(\Rightarrow x^2 - 20x + 112 = 0\)

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.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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:

\(\Rightarrow 2 ( x + {{400} \over {x}}) = 80 \)

\( \Rightarrow 2 ( {{x^2 + 400} \over {x}})\)

\(\Rightarrow 2x^2 + 800 = 80x\)

\(\Rightarrow 2x^2 - 80x + 800 =0\)

\( \Rightarrow x^2 - 40x + 400 = 0\)

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 m².

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 = \({{400} \over {x}} = {{400} \over {20}} = 20 m \)

There are total 24 questions present in ncert solutions for class 10 maths chapter 4 quadratic equations

There are total 20 long question/answers in ncert solutions for class 10 maths chapter 4 quadratic equations

There are total 4 exercise present in ncert solutions for class 10 maths chapter 4 quadratic equations