# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

## Chapter 4 Ex.4.2 Question 1

Find the roots of the following quadratic equations by factorization:

(i) \begin{align} x{}^\text{2}-3x-10=0~\end{align}

(ii) \begin{align}2x{}^\text{2}+x-6=0\end{align}

(iii) \begin{align}\sqrt {\left( 2 \right)} {x^2} + 7x + 5\sqrt 2 \end{align}

(iv) \begin{align}2{x^2} - x + 1/8\end{align}

(v) \begin{align}100x{}^\text{2}-20x+1\end{align}

### Solution

Reasoning:

Roots of the polynomial are same as the zeros of the polynomial. Therefore, roots can be found by factorizing the quadratic equation into two linear factors and equating each to zero.

What is Known?

What is unknown?

(i) \begin{align}\quad x{}^\text{2}-3x-10=0~\end{align}

Steps:

\begin{align} x - 5x + 2x - 10 &= 0 \\ x \left( {x - 5} \right) + 2\left( {x - 5} \right) & = 0 \\ \left( {x - 5} \right)\left( {x + 2} \right) & = 0 \end{align}

\begin{align} x-5 & = 0 \quad \; \rm{and}& x+ 2 &= 0 \\ x &=5 \quad &x &= - 2 \\\end{align}

Roots are $$– 2,\; 5.$$

(ii) \begin{align}\quad2x{}^\text{2}+x-6=0\end{align}

Steps:

\begin{align} 2 x^{2}+x-6 &=0 \\ 2 x^{2}+4 x-3 x-6 &=0 \\ 2 x(x+2)-3(x+2) &=0 \2 x-3)(x+2) &=0 \end{align} \begin{align} 2 x-3 &=0 \qquad \text{and} \qquad x+2=0\\ 2 x &=3 \qquad \text{and} \quad \qquad x=-2 \\ x &={3 \over 2}\;\; \quad \text{and} \qquad x=-2 \end{align} Roots are: \(\begin{align}{3\over2}, - 2\,\end{align}

(iii) \begin{align}\quad\sqrt {\left( 2 \right)} {x^2} + 7x + 5\sqrt 2 \end{align}

Steps:

\begin{align}\sqrt {\left( 2 \right)} x + 7x + 5\sqrt 2 &= 0\\\sqrt {\left( 2 \right)} x + 5x + 2x + 5\sqrt 2 &= 0\\ \begin{bmatrix} \sqrt {\left( 2 \right)} x + 5x \\ + \sqrt {\left( 2 \right)} \times \sqrt {\left( 2 \right)} x \\ + 5\sqrt 2 \end{bmatrix} &= 0\\ \begin{bmatrix} x\left( {\sqrt {\left( 2 \right)} x + 5} \right) \\ + \sqrt 2 \left( {\sqrt {\left( 2 \right)} x + 5} \right) \end{bmatrix} &= 0\\\left( {x\sqrt {\left( 2 \right)} + 5} \right)\left( {x + \sqrt 2 } \right) &= 0 \end{align}

\begin{align} &\left( {\sqrt {\left( 2 \right)} x + 5} \right) = 0\quad \left( {x + \sqrt 2 } \right) = 0\\ & \sqrt {\left( 2 \right)} x = - 5 \quad \qquad x = - \sqrt 2 \\ & x =\!\! \left( \frac{ - 5}{ \sqrt {\left( 2 \right)}} \right) \qquad x = - \sqrt 2 \end{align}

Roots are \begin{align}\frac{ - 5}{ \sqrt {\left( 2 \right)}} , - \sqrt 2 .\end{align}

(iv) \begin{align}\quad2{x^2} - x + 1/8\end{align}

Steps:

Multiplying both sides of the equation by $$8$$

\begin{align} 2(8) x^{2}-8(x)+(8) {1 \over 8} &=0(8) \\ 16 x^{2}-8 x+1 &=0 \\ 16 x^{2}-4 x-4 x+1 &=0 \\ 4 x(4 x-1)-1(4 x-1) &=0 \4 x-1)-1(4 x-1) &=0 \end{align} \begin{align} 4 x-1 &=0 \qquad \qquad 4 x-1 =0 \\ 4 x &=1 \!\qquad \qquad \qquad 4 x =1 \\ x &={1 \over 4} \qquad \qquad \quad \; x ={1 \over 4}\end{align} Roots are \(\begin{align}{1 \over 4},\,\,\,\,{1\over 4}.\end{align}

(v) \begin{align}\quad100x{}^\text{2}-20x+1\end{align}

Steps:

\begin{align} 100 x^{2}-20 x+1 &=0 \\ 100 x^{2}-10 x-10 x+1 &=0 \\ 10 x(10 x-1)-1(10 x-1) &=0 \\ (10 x-1)(10 x-1) &=0 \end{align}

\begin{align} 10 x-1 &=0 \qquad \qquad10 x-1=0 \\ 10 x &=1\qquad \qquad\qquad \!10 x=1 \\ x &=\frac{1}{10} \qquad \qquad\quad x=\frac{1 }{10} \end{align}

Roots are \begin{align} \frac{1}{10},\, \frac{1}{10} \,\,.\end{align}

## Chapter 4 Ex.4.2 Question 2

Solve the problems given in example $$1.$$

(i) John and Jivanti had $$45$$ marbles. Both of them lost $$5$$ marbles each and the product of the no. 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.

### Solution

(i)

What is known?

(a) John and Jivanti together had $$45$$ marbles

(b) Both of them lost $$5$$ marbles each.

(c) Product of number of marbles they now have is $$124.$$

What is Unknown?

Number of marbles John and Jivanti started with (each).

Reasoning:

Let the number of marbles that John had be $$x.$$

The number of marbles Jivanti had will be (Total marbles MINUS the Number of marbles John had) $$= 45-x$$

(a) Both of them lost $$5$$ marbles each:

John $$= x-5$$

Jivanti $$=45 - x - 5 = 40 - x$$

(b) Product of current number of marbles $$=124$$

$\left( {x - 5} \right)\left( {40 - x} \right) = 124$

Steps:

\begin{align}\left( {x - 5} \right)\left( {40 - x} \right) &= 124\\40x - {x^2} - 200 + 5x &= 124\\ - {x^2} + 45x - 200 - 124 &= 0\\ - {x^2} + 45x - 324 &= 0\\{x^2} - 45x + 324 &= 0\\{x^2} - 36x - 9x + 324 &= 0\\x\left( {x - 36} \right) - 9\left( {x - 36} \right) &= 0\\\left( {x - 36} \right)\left( {x - 9} \right) &= 0\end{align}

\begin{align} x - 36 = 0 &\qquad x - 9 = 0\\x = 36 &\qquad x = 9\end{align}

John and Jivanti started with $$36$$ and $$9$$ marbles.

(ii)

What is known?

(a) Cost of each day is $$55\, –$$ (number of toys produced in a day) rupees.

(b) On a particular day the total cost of production was $$\rm{Rs.}\,750.$$

What is Unknown?

Number of toys produced on that day.

Reasoning:

Let the number of toys produced in a day is $$x.$$

(a) Cost of each toy $$= (55 - x)$$ rupees

(b) Total cost of production $$=$$ Cost of each toy $$\times$$ Total number of toys

\begin{align}(55 - x)(x) &= 750\end{align}

Steps:

\begin{align}(55 - x)(x) &= 750\\55x - x &= 750\\55x - x - 750 &= 0\\x - 55x + 750 &= 0\\x - 25x-30x + 750 &= 0\\x(x - 25)-30\left( {x - 25} \right) &= 0\x - 25)(x - 30) &= 0\end{align} \begin{align}x-25 = 0 &\qquad x-30 = 0\\x = 25 &\qquad x = 30\end{align} Number of toys produced on that day is \(25 or $$30.$$

## Chapter 4 Ex.4.2 Question 3

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

### Solution

What is known?

(i) Sum of two numbers is $$27.$$

(ii) Product of two numbers is $$182.$$

What is Unknown?

Two numbers.

Reasoning:

Let one of the numbers be $$x$$. Then the other number will be

$$x +$$ other number $$= 27$$

Other number $$= 27 – x$$

Product of the two numbers $$= 182$$

This can be written in the form of the following equation:

$x\left( {27 - x} \right) = 182$

Steps:

\begin{align}x\left( {27 - x} \right) &= 182\\27x - x &= 182\\27x - x - 182 &= 0\\x - 27x + 182 &= 0\\x - 14x-13x + 182 &= 0\\x\left( {x - 14} \right)-13\left( {x - 14} \right) &= 0\\\left( {x - 13} \right)\left( {x - 14} \right) &= 0\end{align}

\begin{align} x-13 = 0 & \qquad x-14 = 0\\x = 13 & \qquad x = 14\end{align}

The numbers are $$13,\, 14.$$

## Chapter 4 Ex.4.2 Question 4

Find two consecutive positive integers, the sum of whose square is $$365.$$

### Solution

What is known?

Sum of squares of these two consecutive integers is $$365.$$

What is Unknown?

Two consecutive positive integers.

Reasoning:

Let the first integer be $$x.$$

The next consecutive positive integer will be $$x + 1.$$

$x^2+\left( x+1 \right)^2=365$

Steps:

\begin{align}{x^2} + \,{\left( {x + 1} \right)^2} &= 365\\{x^2} + \left( {{x^2} + 2x + 1} \right) &= 365 \\ [\because \left( {a + b} \right)^2= {a^2} + 2ab & + {b^2}] \\2{x^2} + 2x + 1 &= 365\\2{x^2} + 2x + 1 - 365 &= 0\\2{x^2} + 2x - 364 &= 0\\2({x^2} + x - 182)\,& = \,0\\x + x-182 &= 0\\x + 14x-13x-182 &= 0\\x\left( {x + 14} \right)-13\left( {x + 14} \right) &= 0\\ \left( {x - 13} \right)\left( {x + 14} \right) &= 0\end{align}

\begin{align} x - 13&=0 & x + 14 = 0\\x& = 13 & x =- 14\end{align}

Value of $$x$$ cannot be negative (because it is given that the integers are positive).

$\therefore \,\,~x=13 \qquad x\text{ }+\text{ }1\text{ }=\text{ }14$

The two consecutive positive integers are $$13$$ and $$14.$$

## Chapter 4 Ex.4.2 Question 5

The altitude of right triangle is $$7\,\rm{ cm}$$ less than its base. If the hypotenuse is $$13\,\rm{ cm},$$ find the other two sides.

### Solution

What is known?

i) Altitude of right triangle is $$7\, \rm{cm}$$ less than its base.

ii) Hypotenuse is $$13\, \rm{cm}$$

What is Unknown?

The measure of the two sides of a given right triangle.

Reasoning:

In a right triangle, altitude is one of the sides. Let the base be $$x\, \rm{cm.}$$

The altitude will be $$(x - 7)\, \rm{cm.}$$

Next, we can apply the Pythagoras theorem to the given right triangle.

Pythagoras theorem:

$$\text{Hypotenuse}^2= \text{Side 1}^2+ \text{Side 2}^2$$

$13{}^\text{2}\text{ }=\text{ }x{}^\text{2}\text{ }+\text{ }(x\text{ }-7)\text{ }{}^\text{2}$

Steps:

\begin{align}13&= x + {(x - 7)^2}\\169 &= x \!\!+ \!\!x -\!\! 14x \!\!+ \!\!49 \\169 &= 2x - 14x + 49 \end{align}

\begin{align} 2x \!\!- \!\!14x \!\!+ \!49 \!- \!169 & \!=\! 0\\ 2x - 14x \\ - 120 &= 0\\ \frac{{ 2x - 14x - 120 }}{2}&= 0\\x - 7x-60 &= 0\\ x\! - \!12x \!+\! 5x \! - \! 60 &= 0\\ x \! \left( \! x - 12 \! \right) \!+ \! 5\! \left( {x - 12} \right) & = 0\x + 5)(x - 12) &= 0\\ \end{align} \begin{align} x-12 = 0 &\qquad x + 5 = 0\\x = 12 &\qquad \quad\;\; x = - 5\end{align} We know that the value of the base cannot be negative. \(\therefore Base $$= x = 12 \,\rm{cm}$$

Altitude $$= 12 -7 = 5 \,\rm{cm}$$

Lengths of two sides are $$12\, \rm{cm}$$ and $$5\,\rm{ cm.}$$

## Chapter 4 Ex.4.2 Question 6

A cottage industry produces 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 cost of the production on that day is ₹ $$90,$$ find the number of articles produced and the cost of each article.

### Solution

What is known?

(i) On a particular day that cost of production of each article (in rupees) was $$3$$ more than twice the no. of articles produced on that day.

(ii) The total cost of the production is ₹ $$90.$$

What is Unknown?

Number of articles produced and cost of each article.

Reasoning:

Let the number of articles produced on that day be $$x.$$

Therefore, the cost (in rupees) of each article will be $$(3 + 2x)$$

Total cost of production $$=$$ Cost of each article $$\times$$ Total number of articles

$90{\rm{ }} = {\rm{ }}\left( {3{\rm{ }} + {\rm{ }}2x} \right)\left( x \right)$

Steps:

\begin{align}90 &= \left( {3 + 2x} \right)\left( x \right)\\\left( {3 + 2x} \right)\left( x \right) &= 90\\ 3x + 2x^2 &= 90 \end{align}

\begin{align} 2x^2 + 3x-90 &= 0\\ 2x^2 \!\! + \!\! 15x \!- \! 12x \!-90 &= 0\\ \left(\! {2x + 15} \! \right) \! - \! 6 \! \left( 2x \! +\! 15 \right) & = 0\\\left( {2x + 15} \right)\left( {x - 6} \right) &= 0\end{align}

\begin{align} 2x + 15 = 0 &\qquad x-6 = 0\\2x = - 15 &\qquad x = 6\\x = - \left(\frac{15}{2}\right)&\qquad x = 6\end{align}

Number of articles cannot be a negative number.

$\therefore \,\,x = 6$

Cost of each article

\begin{align}& = {\rm{ }}3{\rm{ }} + {\rm{ }}2x\\& = {\rm{ }}3{\rm{ }} + {\rm{ }}2{\rm{ }}\left( 6 \right)\\& = {\rm{ Rs}}{\rm{. }}\,15 \end{align}

Cost of each article is $$\rm{Rs.}\,15.$$

Number of articles produced is $$6.$$