NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

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

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

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.

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.

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.

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.

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.

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.

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)
From (1), (2) and (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.

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.

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

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

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.

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.

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.

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