Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Question 1.

Find the roots of the following quadratic equations if they exist, by the method of completing the square:

(i) 2x^{2} + 7x + 3

(ii) 2x^{2} + x – 4 = 0

(ili) 4x^{2} + 4√3x + 3 = 0

(iv) 2x^{2} + x + 4 = 0

Solution:

(i) Given quadratic equation is

2x^{2} – 7x + 3 = 0

Or 2x^{2} – 7x = -3

Or x^{2} – \(\frac{7}{2}\)x = –\(\frac{3}{2}\)

Or x^{2} – \(\frac{7}{2}\)x + (\(\frac{-7}{4}\))^{2} = \(\frac{-3}{2}+\left(\frac{-7}{4}\right)^{2}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-3}{2}+\frac{49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-24+49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}\)

Or x – \(\frac{7}{4}\) = \(\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}\)

Case I:

When x – \(\frac{7}{4}\) = \(\frac{5}{4}\)

Or x = \(\frac{5}{4}+\frac{7}{4}=\frac{5+7}{4}\)

Or x = \(\frac{12}{4}\) = 3

Case II:

When x – \(\frac{7}{4}\) = \(\frac{-5}{4}\)

Or x = \(\frac{-5}{4}+\frac{7}{4}=\frac{-5+7}{4}\)

Or x = \(\frac{2}{4}=\frac{1}{2}\)

Hence, roots of given quadratic equation is 3, \(\frac{1}{2}\).

(ii) Given Quadratic Equation is

2x^{2} + x – 4 = 0

Or 2x^{2} + x = 4

Or x^{2} + \(\frac{1}{2}\)x = \(\frac{4}{2}\)

Case I:

When x + \(\frac{1}{4}\) = \(\frac{-\sqrt{33}}{4}\)

Or x = \(-\frac{\sqrt{33}}{4}-\frac{1}{4}\)

Or x = \(\frac{-\sqrt{33}-1}{4}\)

Hence, roots of given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\).

(iii) Given quadratic equation is

4x^{2} + 4√3x + 3 = 0

Or 4x^{2} + 4√3x = -3

Or x^{2} + \(\frac{4 \sqrt{3}}{4}\)x = \(\frac{-3}{4}\)

Or x^{2} + √3x = \(\frac{-3}{4}\)

Or x^{2} + √3x + \(\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}\)

Or \(\left(x+\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\frac{3}{4}\)

or (x + \(\frac{\sqrt{3}}{2}\))^{2} = 0

(x + \(\frac{\sqrt{3}}{2}\)) (x + \(\frac{\sqrt{3}}{2}\)) = 0

Either x + \(\frac{\sqrt{3}}{2}\) = 0

x = –\(\frac{\sqrt{3}}{2}\)

Or x + \(\frac{\sqrt{3}}{2}\) = 0

Or x = –\(\frac{\sqrt{3}}{2}\)

Hence, roots of given quadratic equation are –\(\frac{\sqrt{3}}{2}\) and –\(\frac{\sqrt{3}}{2}\).

(iv) Given quadratic equation is

2x^{2} + x + 4 = 0

2x^{2} + x = -4

x^{2} + \(\frac{1}{2}\)x = \(-\frac{4}{2}\)

Or x^{2} + \(\frac{1}{2}\)x + (\(\frac{1}{4}\))^{2} = -2 + (\(\frac{1}{4}\))^{2}

Or \(\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}\)

Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}\)

Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}<0\)

∴ square of any number cannot be negative. So, (x + \(\frac{1}{4}\))^{2} cannot be negative for any real x.

∴ There is no real x whith satisfied the given quadratic equation.

Hence, given quadratic equation has no real roots.

Question 2.

Find the roots of the quadratic equations given in Q. 1 by applying the quadratic formula. Which of the above two

methods do you prefer, and why?

Solution:

(i) Given quadratic equation is

2x^{2} – 7x + 3 = 0

Compare it with ax^{2} + bx + c = 0

a = 2, b = -7, c = 3

Now, b^{2} – 4ac = (-7)^{2} 4 x 2 x 3

= 49 – 24

= 25 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-7) \pm \sqrt{25}}{2 \times 2}=\frac{7 \pm 5}{4}\)

= \(\frac{7+5}{4} \text { and } \frac{7-5}{4}\)

= \(\frac{12}{4} \text { and } \frac{2}{4}\)

= 3 and \(\frac{1}{2}\)

Hence, 3 and \(\frac{1}{2}\) are the roots of given quadratic equation.

(ii) Given quadratic equation is

2x^{2} + x – 4 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 2, b = 1, c = -4

Now,

b^{2} – 4ac = (1)^{2} – 4 × 2 × 4

= 1 + 32 = 33 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-1 \pm \sqrt{33}}{2 \times 2}=\frac{-1 \pm \sqrt{33}}{4}\)

= \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

Hence, \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\) are the roots of given quadratic equation.

(iii) Given quadratic equation is

4x^{2} + 4√3x + 3 = 0

Compare it with ax^{2} + bx + c = 0

a = 4, b = 4√3, c = 3

b^{2} – 4ac = (4√3)^{2} – 4 × 4 × (3)

= 48 – 48 = 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2 \times 4}\)

= \(-\frac{4 \sqrt{3}}{8}\), \(-\frac{4 \sqrt{3}}{8}\)

= –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\)

Hence, –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\) are the roots of given quadratic equation.

(iv) Given quadratic equation is 2x^{2} + x + 4 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 2, b = 1, c = 4

Now, b^{2} – 4ac = (1)^{2} – 4 × 2 × 4

= 1 – 32 = -31 < 0

But

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Since the square of a real number cannot be negative, therefore x will not have any real value.

Hence, there are no real roots for the given quadratic equation.

From above two questions, we used two methods to find the roots of the quadratic equations. It is very clear from above discussion that quadratic formula method is very convenient as compared to method of completing the square.

Question 3.

Find the roots of the following equations:

(i) x – \(\frac{1}{x}\) = 3, x ≠ 0

(ii) \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ -4, 7

Solution:

(i) Given Equation is

x – \(\frac{1}{x}\) = 3

Or \(\frac{x^{2}-1}{x}\) = 3

Or x^{2} – 1 = 3x

Or x^{2} – 3x – 1 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 1, b = -3, c = -1

Now, b^{2} – 4ac = (-3)^{2} – 4 . 1 . (-1)

= 9 + 4 = 13 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-3) \pm \sqrt{13}}{2 \times 1}\)

= \(\frac{3 \pm \sqrt{13}}{2 \times 1}\)

= \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\)

Hence, \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\) are the roots of given quadratic equation.

(ii) Given equation is

-11 × 30 = 11 (x2 – 3x – 28)

Or -30 = x^{2} – 3x – 28

Or x^{2} – 3x – 28 + 30 = 0

Or x^{2} – 3x + 2 = 0

Compare it with ax^{2} + bx + c = 0

a = 1, b = – 3, c = 2

Now, b^{2} – 4ac = (-3)^{2} – 4 × 1 ×2

= 9 – 8 = 1 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-3) \pm \sqrt{1}}{2 \times 1}=\frac{3 \pm 1}{2}\)

= \(\frac{3+1}{2}\) and \(\frac{3+1}{2}\)

= \(\frac{4}{2}\) and\(\frac{2}{2}\) and 1

Hence, 2 and 1 are the roots of given quadratic equation.

Question 4.

The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). F1nd his present age.

Solution:

Let Rehman’s present age = x years

3 years ago Rehman’s age (x – 3) years

5 years from now Rehman’s age =(x + 5) years

According to question,

\(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

Or \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+5 x-3 x-15}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+2 x-15}=\frac{1}{3}\)

Or 6x + 6 = x^{2} + 2 -15

Or x^{2} + 2x – 15 – 6x – 6 = 0

Or x^{2} – 4x – 21 = 0, which is quadratic in x.

So compare it with ax^{2} + bx + c =0

a = 1, b = -4, c = -21

Now, b^{2} – 4ac = (- 4)^{2} 4 × 1 × (-21)

= 16 + 84 = 100 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

x = \(\frac{-(-4) \pm \sqrt{100}}{2 \times 1}\)

= \(\frac{4 \pm 10}{2}\)

= \(\frac{4+10}{2}\) and \(\frac{4-10}{2}\)

\(\frac{14}{2}\) and \(\frac{-6}{2}\)

= 7 and -3

∵ age cannot be negative,

so, we reject x = – 3

∴ x = 7

Hence, Rehman’s present age = 7 years.

Question 5.

In a class test, the sum of Shefall’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less ¡n English, the product of their marks would have been 210. Find her marks In the two subjects.

Solution:

Let Shefali get marks in Mathematics = x

Shefali’s marks in English = 30 – x

According to 1st condition,

Shefali’s marks in Mathematics = x + 2

and Shefali’s marks in English = 30 – x – 3 = 27 – x

∴ Their product = (x + 2) (27 – x)

= 27x – x^{2} + 54 – 2x

= x^{2} + 25x + 54

According to 2nd condition,

-x^{2}+ 25x+ 54 = 210

Or -x^{2} + 25x + 54 – 210 = 0

Or -x^{2} + 25x – 156 = 0

Or x^{2} – 25x+ 156 = o

Compare it with ax^{2} + bx + c = O

a = 1, b = -25, c = 156

Now, b^{2} – 4ac = (-25)^{2} – 4 × 1 × 156

= 625 – 624 = 1 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-25) \pm \sqrt{1}}{2 \times 1}\)

= \(\frac{25 \pm 1}{2}\)

= \(\frac{25+1}{2}\) and \(\frac{25-1}{2}\)

= \(\frac{26}{2}\) and \(\frac{24}{2}\)

= 13 and 12.

Case I:

When x = 13

then Shefaiis marks in Maths = 13

Shefali’s marks in English = 30 – 13 = 17.

Case II:

When x = 12

then Shefalis marks in Maths = 12

Shefali’s marks in English = 30 – 12 =18.

Hence, Shefalis marks in two subjects are 13 and 17 Or 12 and 18.

Question 6.

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.

Solution:

Let shorter side of rectangular field = AD = x m

Longer side of rectangular field = AB = (x + 30) m

and diagonal of rectangular field = DB = (x + 60) m

In rectangle. the angle between the length and breadth is right angle.

∴ ∠DAB = 90°

Now, in right angled triangle DAB, using Pythagoras Theorem,

(DB)^{2} = (AD)^{2} + (AB)^{2}

(x + 60)^{2} = (x)^{2} + (x + 30)^{2}

Or x^{2} + 3600 + 120x = x^{2} + x^{2} + 900 + 60x

Or x^{2} + 3600 + 120x – 2x^{2} – 900 – 60x = 0

Or -x^{2} + 60x + 2700 = 0

Or x^{2} – 60x – 2700 = 0

Compare it with ax^{2} + bx + e = O

∴ a = 1, b = -60, c = -2700

and b^{2} – 4ac = (-60)^{2} – 4. 1 . (-2700)

= 3600 + 10800 = 14400 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-60) \pm \sqrt{14400}}{2 \times 1}\)

= \(\frac{60 \pm 120}{2}\)

= \(\frac{60+120}{2}\) and \(\frac{60-120}{2}\)

= \(\frac{180}{2}\) and \(\frac{-60}{2}\)

= 90 and – 30

∴ length of any side cannot be negative

So, we reject x = -30

∴ x = 90

Hence, shorter side of rectangular field = 90 m

Longer side of rectangular field = (90 + 30) m = 120 m.

Question 7.

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.

Solution:

Let larger number = x .

Smaller number = y

According to 1st condition,

x^{2} – y^{2} = 180 ……………(1)

According to 2nd condition,

y^{2} = 8x

From (1) and (2), we get

x^{2} – 8x = 180

Or x^{2} – 8x – 180 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = -1, b = -8, c = -180

and b^{2} – 4ac = (-8)^{2} – 4 × 1 × (-180)

= 64 + 720 = 784 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-8) \pm \sqrt{784}}{2 \times 1}\)

= \(\frac{8 \pm 28}{2}\)

= \(\frac{8+28}{2}\) and \(\frac{8-28}{2}\)

= \(\frac{36}{2}\) and \(\frac{-20}{2}\)

= 18 and -10

When x = – 10 then from (2),

y^{2} = 8 (- 10) = – 80, which is impossible.

So, we reject x = – 10

When x = 18 then from (2).

y^{2} = 8(18) = 144

Or y = ±√144

Or y = ± 12

Hence, required numbers are 18 and 12 Or 18 and -12.

Question 8.

A train travels 360 km ¡t a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let constant speed of the train = x km/hour

Distance covered by the train = 360 km

Time taken by the train = \(=\frac{\text { distance }}{\text { speed }}\)

(∵ speed = \(\frac{\text { Distance }}{\text { Time }}\))

= \(\frac{360}{x}\)

Increased speed of the train = (x + 5) km/hour

∴ Time taken by the train with increased speed = \(\frac{360}{x+5}\) hour

According to question

Or 1800 = x^{2} + 5x

Or x^{2} + 5x – 1800 = 0

Compare it with, ax^{2} + bx + c = 0

a = 1, b = 5, c = – 1800

and b^{2} – 4ac = (5)^{2} 4 × 1 × (- 1800)

= 25 + 7200 = 7225 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-5 \pm \sqrt{7225}}{2 \times 1}\)

= \(\frac{-5 \pm 85}{2}\)

= \(\frac{-5+85}{2}\) and \(\frac{-5-85}{2}\)

= \(\frac{80}{2}\) and \(\frac{-90}{2}\)

= 40 and – 45

∵ speed of any train cannot be negative.

So, we reject x = – 45

x = 40

Hence, speed of train = 40 km/hour.

Question 9.

Two water taps together can fill a tank in 9\(\frac{3}{8}\) 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.

Solution:

Let time taken by larger tap to fill the tank = x hours.

Time taken by smaller tap to fill the tank = (x + 10) hours

In case of one hour:

Larger tap can fill the tank = \(\frac{1}{x}\)

Smaller tap can fill the tank = \(\frac{1}{x+10}\)

∴ Larger and smaller tap fill the tank = \(\frac{1}{x}\) + \(\frac{1}{x+10}\) ………….(1)

But, two taps together can fill the tank = 9\(\frac{3}{8}\)hour = \(\frac{75}{8}\) hour

Now, two taps together can fill the tank in one hour = \(\frac{8}{75}\) ……………..(2)

From (1) and (2), we get

\(\frac{1}{x}+\frac{1}{x+10}=\frac{8}{75} \)

Or \(\frac{x+10+x}{x(x+10)}=\frac{8}{75}\)

Or \(\frac{2 x+10}{x^{2}+10 x}=\frac{8}{75}\)

Or 75(2x + 10) = 8(x^{2} + 10x)

Or 150x + 750 = 8x^{2} + 80x

Or 8x^{2} + 80x – 150x – 750 = 0

Or 8x^{2} – 70x – 750 = 0

Or 4x^{2} – 35x – 375 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 4, b = -35, c = -375

and b^{2} – 4ac = (35)^{2} -4 × 4 × (-375)

= 1225 + 6000 = 7225 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-35) \pm \sqrt{7225}}{2 \times 4}\)

= \(\frac{35 \pm 85}{8}\)

= \(\frac{35+85}{8}\) and \(\frac{35-85}{8}\)

= \(\frac{120}{8}\) and \(\frac{-50}{8}\)

= 15 and \(\frac{-25}{4}\)

∵ time cannot be negative.

So,we reject x = \(\frac{-25}{4}\)

∴ x = 15

Hence, larger water tap fills the tank = 15 hours

and smaller water tap fills the tank = (15 + 10) hours = 25 hours.

Question 10.

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 to intermediate stations), if the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.

Solution:

Let average speed of passenger train = x km/hour

Average speed of express train = (x+ 11) km/hour

Distance between Mysore and Bangalore = 132 km

Time taken by passenger train = \(\frac{132}{x}\) hour

[∵ Speed = \(=\frac{\text { Distance }}{\text { Time }}\) ]

Time taken by express train‚ = \(\frac{132}{x+11}\) hour

According to question,

Or 1452 = x^{2} + 11x

Or x^{2} + 11x – 1452 = 0

Compare it with ax^{2} + bx + c = 0

∴ a = 1, b = 11, c = -1452

and b^{2} – 4ac = (11)^{2} – 4 × 1 × (- 1452)

= 121 + 5808 = 5929 > 0

∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-11 \pm \sqrt{5929}}{2 \times 1}\)

= \(\frac{-11 \pm 77}{2}\)

= \(\frac{-11+77}{2}\) and \(\frac{-11-77}{2}\)

= \(\frac{66}{2}\) and \(\frac{-88}{2}\) = 33 and -44

∵ speed of any train cannot be negative

∴ x = 33

Hence, speed of passenger train = 33 km/hour

and speed of express train = (33 + 11) km/hour = 44 km/hour.

Question 11.

Sum of the areas of two squares is 468 m^{2}. If the difference of their perimeters

is 24 m, find the sides of the two squares.

Solution:

In case of larger square

Let length of each side of square = x m

Area of square = x^{2} m^{2}

Perimeter of square = 4x m

In case of smaller square:

Let lenth of each side of square = y m

Area of square = y^{2} m^{2}

Perimeter of square = 4y m

According to 1st condition,

x^{2} + y^{2} = 468 …………….(1)

According to 2nd condition,

4x – 4y = 24

Or 4(x – y) = 24

Or x – y = 6

x = 6 + y

From (1) and (2), we get

(6 + y)^{2} + y^{2} = 468

Or 36 + y^{2} + 12y + y^{2} = 468

Or 2y^{2} + 12y + 36 – 468 = 0

Or 2y^{2} + 12y – 432 = 0

Or y^{2} + 6y – 216 = 0

Compare it with ay^{2} + by + c = 0

∴ a = 1, b = 6, c = -216

and b^{2} – 4ac = (6)^{2} – 4 × 1 × (- 216) = 36 + 864 = 900 > 0

∴ y = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-6 \pm \sqrt{900}}{2 \times 1}\)

= \(\frac{-6 \pm 30}{2}\)

= \(\frac{-6+30}{2}\) and \(\frac{-6-30}{2}\)

= \(\frac{24}{2}\) and \(\frac{-36}{2}\) = 12 and -18

∵ length of square cannot be negative

So, we reject y = – 18

∴ y = 12

From (2), x = 6 + 12 = 18

Hence, sides of two squares are 12 m and 18 m.