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

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

## PSEB 10th Class Maths Solutions 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:

Question 4.

The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 13. 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,

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

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

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

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.

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.

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

∵ 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.

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