# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.4

Quadratic Equations

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

**Video 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

**Video 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

**Video 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

**Video 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

**Video 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 \( = 20\,\rm{m.}\)

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