# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1

## Chapter 4 Ex.4.1 Question 1

Check whether the followings are quadratic equations:

i)   \begin{align}{(x + 1)^2} = 2\left( {x - 3} \right)\end{align}

ii)  \begin{align} \left( {{x^2} - 2x} \right) = \left( { - 2} \right)\left( {3 - x} \right)\end{align}

iii) \begin{align} \begin{bmatrix}\left( {x - 2} \right)\left( {x + 1} \right) \\ = \left( {x - 1} \right)\left( {x + 3} \right) \end{bmatrix}\end{align}

iv) \begin{align} \left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)\end{align}

v)  \begin{align} \begin{bmatrix} \left( {2x - 1} \right)\left( {x - 3} \right) \\ = \left( {x + 5} \right)\left( {x - 1} \right) \end{bmatrix} \end{align}

vi) \begin{align} {x^2} + 3x + 1 = {\left( {x - 2} \right)^2}\end{align}

vii)  \begin{align} {\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)\end{align}

viii) \begin{align} {x^3} - 4x - x + 1 = {\left( {x - 2} \right)^3}\end{align}

### Solution

Reasoning:

Standard form of a quadratic equation is \begin{align}a{x^2} + bx + c = 0\end{align} where \begin{align}a \ne 0\end{align} and degree of quadratic equation is $$2.$$

What is unknown?

To check whether the given equation is a quadratic equation

Steps:

i)

\begin{align}\quad{(x + 1)^2} = 2\left( {x - 3} \right)\end{align}

\begin{align}&{\left( {{{a + b}}} \right)^{{2}}} = {{{a}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ + 2ab}}\\&{{{x}}^{{2}}}{{ + 2x + 1}} = \left( {{{2x - 6}}} \right)\\&{{{x}}^{{2}}}{{ + 2x + 1 - }}\left( {{{2x - 6}}} \right) = 0\\&x^2+2x+1-2x+6\\&{{{x}}^{{2}}}{{ + 7}} = 0\end{align}

Here degree of \begin{align}x^2 + 7 = 0\end{align}is $$2.$$

$$\therefore$$  It is a quadratic equation.

ii)

\begin{align}\quad\left( {{x^2} - 2x} \right) = \left( { - 2} \right)\left( {3 - x} \right)\end{align}

\begin{align}{x^2} - 2x &= - 6 + 2x\\{x^2}\, - 2x - 2x + 6 &= 0\\{x^2} - 4x + 6 &= 0\end{align}

Degree $$= 2.$$

$$\therefore$$  It is a quadratic equation.

iii)

\begin{align}\quad\left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)\end{align}

\begin{align}\begin{bmatrix}{x^2} - 2x + \\x - 2 \end{bmatrix}&= {x^2} + 3x - x - 3 \\{x^2} - x - 2 &= {x^2} + 2x - 3\\\begin{bmatrix}{x^2} - x - 2 - \\{x^2} - 2x + 3\end{bmatrix} &= 0\\ - 3x + 1 &= 0\end{align}

Degree $$= 1$$

$$\therefore$$ It is not a quadratic equation.

iv)

\begin{align}\quad\left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)\end{align}

\begin{align}&2{x^2} + x - 6x - 3 = {x^2} + 5x\\&2{x^2} - 5x - 3 = {x^2} + 5x\\&2{x^2} - 5x - 3 - {x^2} - 5x = 0\\&{x^2} - 10x - 3 = 0\end{align}

Degree $$= 2$$

$$\therefore$$ It is a quadratic equation.

v)

\begin{align}\quad\left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)\end{align}

\begin{align}\begin{bmatrix}2{x^2} \! - \! 6x \! -\\ x \! + \! 3\end{bmatrix} & \! = \! {x^2} \! - \! x \! + \! 5x \! - \! 5 \\2{x^2} \! - \! 7x \! + \! 3 & \! = \! {x^2} \! + \! 4x \! - \! 5\\\begin{bmatrix}2{x^2} \! - \! 7x \! + \! 3 \! -\\ {x^2} \! - \! 4x \! + \! 5 \end{bmatrix}& \! = \! 0\\{x^2} \! - \! 11x \! + \! 8 & \! = \! 0\end{align}

Degree $$= 2$$

$$\therefore$$ It is a quadratic equation

vi)

\begin{align}{}\quad{x^2} + 3x + 1 = {\left( {x - 2} \right)^2}\end{align}

\begin{align} {x^2} + 3x + 1 &= {x^2} - 4x + 4 \\ [\because ( a - b )^2 &= {a^2} - 2ab + {b^2} ] \\ \begin{bmatrix} {x^2} + 3x + 1 - \\{x^2} + 4x - 4\end{bmatrix} &= 0 \\ 7x - 3 &= 0\end{align}

Degree $$= 1$$

$$\therefore$$ It is not a quadratic equation.

vii)

\begin{align}\quad{\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)\end{align}

Since,

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

\begin{align} & \begin{bmatrix} {x^3} + 3{x^2}\left( 2 \right) +\\ 3{\left( 2 \right)^2}x + {\left( 2 \right)^3} \end{bmatrix}= 2{x^3} - 2x \\& \begin{bmatrix} {x^3} + 6{x^2} + 12x +\\ 8 - 2{x^3} + 2x \end{bmatrix} = 0\\& - {x^3} + 6{x^2} + 14x + 8 = 0\end{align}

Degree $$= 3$$

$$\therefore$$ It is not a quadratic equation.

viii)

\begin{align}\quad{x^3} - 4x - x + 1 = {\left( {x - 2} \right)^3}\end{align}

\begin{align} \begin{bmatrix} {x^3} \! - \! 4{x^2} \! - \! \\ x \! + \! 1 \end{bmatrix} & \! = \! \begin{bmatrix} {x^3} \! - \! 3 {\left( x \right)^2}\left( 2 \right) \! + \! \\ 3\left( x \right){\left( 2 \right)^2} \! - \! {\left( 2 \right)^3} \end{bmatrix} \\ \because \, \left( a \! - \! b \right)^3 & \! = \! {a^3} \! - \! 3{a^2}b \! + \! 3a{b^2} \! - \! {b^3} \\ \begin{bmatrix} {x^3} \! - \! 4{x^2} \! -\\ x \! + \! 1\end{bmatrix} & \! = \! \begin{bmatrix} {x^3} \! - \! 6{x^2} \! + \! \\ 12x \! - \! 8 \end{bmatrix} \\ \begin{bmatrix}{x^3} \! - \! 4{x^2} \! - \! x \! +\\ 1 \! - \! {x^3} \! + \! 6{x^2} \! - \! \\ 12x \! + \! 8 \end{bmatrix} & \! = \! 0\\ 2{x^2} \! - \! 13x \! + \! 9 & \! = \! 0 \end{align}

Degree $$= 2$$

$$\therefore$$ It is a quadratic equation

## Chapter 4 Ex.4.1 Question 2

Represent the following situation in the form of quadratic equations.

(i) The area of a rectangular plot is $$528\,\rm{m}^2.$$ 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.

(ii) The product of two consecutive positive integers 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) $$3$$ years from now will be $$360.$$ We would like to find Rohan’s present age.

(iv) A train travels a distance of $$480 \,\rm{km}$$ at a uniform speed. If the speed had been $$8 \,\rm{km/h}$$ less, then it would have taken $$3$$ hours more to cover the same distance. We need to find the speed of the train.

### Solution

(i)

What is Known?

(a) Area of a rectangular plot is $$528\, \rm{m}^2.$$

(b) Length of the plot (in meters) is one more twice its breadth.

What is unknown?

Quadratic equation for the situation

Reasoning:

We know that the area of a rectangle can be expressed as the product of its length and breadth. Since we don’t know the length and breadth of the given rectangle, we assume the breadth of the plot to be a variable ($$x$$ meters). Then, we use the given relationship between the length and breadth: length $$= 1 + 2$$ times breadth.

Therefore, Length $$= 1 + 2x$$
Area of rectangle $$=$$ Length $$\times$$ Breadth

Steps:

Breadth $$= x$$

Length $$= 2x + 1$$

Area of rectangular plot $$=$$ Length $$\times$$ Breadth$$= 528\;\rm{m}^2$$

\begin{align}\left( {2x + {{ }}1} \right) \times \left( x \right) &= {{ }}528\\ 2x{{ }} + {{ }}x &= {{ }}528\\2x{{ }} + x-{{ }}528 &= {{ }}0 \end{align}

Quadratic equation is $$2x + x-528 = 0,$$ where $$x$$ is the breadth of the rectangular plot.

(ii)

What is Known?

Product of two consecutive integers is $$306.$$

What is unknown?

Quadratic equation for the situation.

Reasoning:

Let the first integer be $$x.$$ Since the integers are consecutive, the next positive integer is $$x +1.$$

We know that:

First integer $$\times$$ Next integer $$= 306$$

Therefore,

\begin{align}x( {x + {\text{ }}1} )=306\end{align}

Steps:

\begin{align}x(x + 1)&=306\\x + x &= 306\\x + x- 306&=0\end{align}

Quadratic equation is $$x^2 + x-{\rm{ }}306{\rm{ }} = {\rm{ }}0.$$

(Where $$x$$ is the first integer)

(iii)

What is Known?

(a) Rohan’s mother is $$26$$ years older than him.

(b) Product of their ages $$3$$ year from now will be $$360.$$

What is the unknown?

Quadratic equation for the known situation.

Reasoning:

Let assume that Rohan’s present age is $$x$$ years. Then, from first condition, mother’s age is ($$x + 26$$) years. Three years from now, Rohan’s age will be $$x +3$$ and, Rohan’s mother age will be $$x +3+ 26 = x + 29.$$ The product of their ages is $$360.$$

\begin{align}( x + 3) \times (x + 29) = 360\end{align}

Steps:

\begin{align}(x+3) \times (x+29)&=360 \\x^2+29x+3x+87&=360 \\x^2+32x+87-360&=0 \\ x^2+32x-273&=0 \end{align}

Quadratic equation is \begin{align}x^{2}+32x-273=0.\end{align}

Where $$x$$ is the present age of Rohan’s age.

(iv)

What is Known?

(a) Distance covered is $$480\,\rm{ km}$$ (at a uniform speed).

(b) If the speed had been $$8\,\rm{ km /h}$$ less, then it would have taken $$3$$ hours more to cover the same distance.

What is unknown?

Quadratic equation to find the speed of the train.

Reasoning:

Distance is equal to speed multiplied by time. Let the speed be $$\text{s km/h}$$ and time be $$t$$ hours.

$$D = \rm{st}$$

$$480 = \rm{st}$$

As per known the given conditions, for the same distance covered at a speed reduced by 8 km /h, the time taken would have increased by 3 hours. Therefore:

$\left( {s-8} \right)\left( {t + 3} \right) = 480 \qquad \left( \rm{i} \right)$

Steps:

\begin{align}(s-8)(t+3) &=480 \\ s t+3 s-8 t-24 &=480 \\ 480+3 s-8(480 / s)-24 &=480 \\ 3 s-3840 / s-24 &=0 \\ 3 s^{2}-3840 -24(s)&=0 \\ 3 s^{2}-24 s-3840 &=0 \\ \frac{3 s^{2}-24 s-3840}{3} &=0 \\ s^{2}-8 s-1280 &=0 \end{align}

\begin{align}{S^2} - 8S-1280 = 0\end{align} is the quadratic equation, where $$\rm{s}$$ is the speed of the train .

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Ncert Class 10 Exercise 4.1
Ncert Solutions For Class 10 Maths Chapter 4 Exercise 4.1
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