# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.4

## Question 1

Find the nature of the roots of the following quadratic equations, if the real root exists, find them.

(i) $$2x^2-3x+5=0$$

(ii) $$3x^2 - 4\sqrt {\left( 3 \right)} x + 4 = 0$$

(iii) $$2x^{2}-6x+3=0$$

### Solution

What is Unknown?

Nature of roots.

Reasoning:

The general form of a quadratic equation is $$ax^{2}+bx+c=0$$

$$b^{2}{ }-{ }4ac$$ is called the discriminant of the quadratic equation and we can decide whether the real roots are exist or not based on the value of the discriminant:

i) Two distinct real roots, if $$b^{2}{ }-{ }4ac{ }>{ }0$$

ii) Two equal real roots, if $$b^{2}{ }-{ }4ac{ }={ }0$$

iii) No real roots, if $$b^{2}{ }-{ }4ac{ }<{ }0$$

(i) $$2x^2-3x+5=0$$

Steps:

(i) $$2x^2-3x+5=0$$

$$a = 2, \;b = -3\; c = 5$$

\begin{align}{b^2} - 4ac &= {{( - 3)}^2} - 4(2)(5)\\&= 9 - 40\\&= - 31|\\{b^2} - 4ac &> 0\end{align}

$$\therefore$$ No real roots.

(ii) $$3x^2 - 4\sqrt {\left( 3 \right)} x + 4 = 0$$

$$a = 3,\;b = - 4\sqrt ( 3),\;c = 4$$

\begin{align}{b^2} - 4ac &= {\left( { - 4\sqrt {(3} )} \right)^2} - 4(3)(4)\\&= 16 \times 3 - 4 \times 4 \times 3\\&= (16 \times 3) - (16 \times 3)\\&= 0\\{b^2} - 4ac &= 0\end{align}

$$\therefore\;$$Two equal real roots.

\begin{align}{\text{ Sum of roots }} &= 2x{ = - \frac{b}{a}}\\x& = - \frac{b}{{2a}}\\x &= \frac{{ - ( - 4\sqrt 3 )}}{{2(3)}}\\&= \frac{{4\sqrt 3 }}{{2\sqrt 3 \sqrt 3 }}\\&= \frac{2}{{\sqrt 3 }}\end{align}

Roots are \begin{align} \frac{2}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}. \end{align}

(iii) $$2x^{2}-6x+3=0$$

$$a = 2,\; b = -6, \;c = 3$$

\begin{align}{b^2} - 4ac &= {{( - 6)}^2} - 4(2)(3)\\&= 36 - 24\\&= 12|\\{b^2} - 4ac &> 0\end{align}

$$\therefore\;$$Two distinct real roots.

\begin{align}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ x &= \frac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)(3)} }}{{2(2)}}\\&= \frac{{6 \pm \sqrt {12} }}{4}\\ x &= \frac{{6 \pm 2\sqrt 3 }}{4}\\x &= \frac{{3 \pm \sqrt 3 }}{2}\end{align}

Roots are \begin{align}x = \frac{{3 + \sqrt 3 }}{2},\frac{{3 - \sqrt 3 }}{2}.\end{align}

## Question 2

Find the value of $$k$$ for each of the following quadratic equation , so that they have two equal roots.

(i) $$2x^\text{2}+kx+3=0$$

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

### Solution

What is known?

value of $$k.$$

What is Known?

Quadratic equation has equal real roots.

Reasoning:

Since the quadratic equation has equal real roots:

Discriminant $$b^\text{2}-4ac=0$$

Steps:

(i) $$2x^\text{2}+kx+3=0$$

$a= 2,\;b = k,\;c = 3$

\begin{align}{b^2} - 4ac &= 0\\{{(k)}^2} - 4(2)(3) &= 0\\{k^2} - 24 &= 0\\{k^2} &= 24\\k &= \sqrt {24} \\k &= \pm \sqrt {2 \times 2 \times 2 \times 3} \\k& = \pm 2\sqrt 6 \end{align}

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

$a = k,\;b = - 2k,\;c = 6$

\begin{align}{b^2} - 4ac &= 0\\{{( - 2k)}^2} - 4(k)(6) &= 0\\4{k^2} - 24k &= 0\\4k(k - 6) &= 0\\k = 6 & \qquad k = 0\\\end{align}
If we consider the value of $$k$$ as $$0,$$ then the equation will not longer be quadratic.

Therefore, $$k = 6$$

## Question 3

Is it possible to design a rectangular mango groove whose length is twice its breadth, and the area is $$800\,\rm{m}^².$$ If so, find its length and breadth.

### Solution

What is known?

i) Mango groove length is twice its breadth.

ii) Area $$=800\,\rm{m}^².$$

What is Unknown?

Finding the possibility of mango groove and if possible length and breadth.

Reasoning:

Let the breadth of rectangle $$x\,\rm{ m.}$$

$$\text {Length}= 2x\,\rm{m}$$

$$\text{Area}= \rm{Length} × \rm{Breadth}$$

Steps:

$$\text{Area}=\rm{ Length} × \rm{Breadth}$$

\begin{align}800 &= x \times 2x\\2{x^2} &= 800\\{x^2} &= \frac{{800}}{2}\\{x^2} &= 400\\ {x^2} - 400 &= 0\end{align}

Discriminant of a quadratic equation is $$b² - 4ac.$$ Comparing $${x^2} - 400 = 0$$ with $$ax^{2}+bx+c=0$$:

$$a = 1,\, b = 0, \,c = - 400$$

\begin{align}{b^2} - 4ac &= {{(0)}^2} - 4(1)( - 400)|\\ &= + 1600 > 0 \end{align}

$$\therefore$$ Yes, it is possible to design a mango groove.

\begin{align}{x^2} - 400 &= 0\\{x^2} &= 400\\x\, &= \pm 20\end{align}

Value of $$x$$ can’t be negative value as it represents the breadth of the rectangle.

$$\therefore\; x = 20\,\rm{m}$$

So, yes, possible to design the mango groove

Length $$= 2x = 2(20) = 40\,\rm{m}$$

Breadth $$= x = 20\,\rm{m}$$

## Question 4

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends are $$20$$ years. Four years ago, the product of their age in year was $$48$$ years.

### Solution

What is Known?

i) The sum of the ages of two friends are $$20$$ years.

ii) Four years ago, the product of their age in year was $$48$$ years.

What is UnKnown?

Checking the possibility of the situation and if yes, find the present ages.

Reasoning:

Let the age of friend $$1$$ be $$x$$ years.

Then,

i) Age of friend $$2 = 20 \,–$$ age of friend $$1= 20 - x$$

ii) Four years ago, age of friend $$1 = x – 4$$

Four years ago, age of friend $$2 = 20 - x - 4$$

Product of their ages:

$$(x - 4)(20 - x - {\rm{ }}4) = 48$$

Steps:

\begin{align}(x - 4)(16 - x)& = 48\\16x - {x^2} - 64 + 4x &= 48\\ - {x^2} + 20x - 64 &= 48\\ - {x^2} + 20x - 64 - 48&= 0\\- {x^2} + 20x - 112 &= 0\\{x^2} - 20x + 112 &= 0\end{align}

Let’s find the discriminant: $$b² - 4ac$$

$$a = 1,\, b = -20,\, c = 112$$

\begin{align}b^2- 4ac& = ( - 20)^2 - 4(1)(112)\\& = 400 - 448\\& = - 48\\b^2 - 4{ac} &< 0\end{align}

Therefore, there are no real roots. So, this situation is not possible.

## Question 5

Is it possible to design a rectangular park of perimeter of $$80\,\rm{m}$$ and area $$400\,\rm{m}^2$$?

If so, find its length and breadth.

### Solution

What is Known?

Perimeter of rectangular park $$= 80\,{\rm{ m}}$$

Area of rectangle $$=400\, \rm{m}^2$$

What is Unknown?

Checking the possibility to design a rectangular park with the given condition , If yes find its perimeter .

Reasoning:

Perimeter of rectangle

$$= 2(l + b) = 80 \; \ldots (1)$$

Area of rectangle $$= lb = 400 \; \ldots \left( 2 \right)$$

We will use the $$1^\rm{st}$$ equation to express $$l$$ in the form of $$b.$$ Then, we will substitute this value of $$l$$ in equation $$2.$$

Steps:

\begin{align}2(l + b) &= 80\l + b) &= 40\\l &= 40 - b\end{align} Substituting the value of \(l = 40 - b in equation (2)

$\begin{array}{*{20}{l}} {(40 - b)(b) = 400}\\ {40b - {b^2} = 400}\\ {40b - {b^2} - 400 = 0}\\ {{b^2} - 40b + 400 = 0}\\ {} \end{array}$

Let’s find the discriminant: $$b² - 4ac$$

\begin{align}a &= 1,\;b= - 40,\;c = 400\\b^2 - 4ac& = {{( - 40)}^2} - 4(1)(400)\\& = 1600 - 1600\\& = 0\end{align}

Therefore, it is possible to design a rectangular park with the given condition:

\begin{align}x& = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - b \pm 0}}{{2a}}\\& = \frac{{ - ( - 40)}}{{2(1)}}\\&= \frac{{40}}{2}\\&= 20\end{align}

$$b = 20,\, l = 40-b = 20$$

$$\therefore\;$$Yes, possible to design a rectangular park with side $$= 20\,\rm{m.}$$