Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.3 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.3

Question 1.

Find the roots of the following quadratic equations, if they exist, by the method of completing the square. (A.P.Mar. 16, 15)

i) 2x^{2} + x – 4 = 0

Solution:

Given : 2x^{2} + x – 4 = 0

⇒ 2x^{2} + x = 4

⇒ (\(\sqrt{2}\))^{2} + x = 4

⇒ (\(\sqrt{2}\)x)^{2} + 2.\(\sqrt{2}\). x. \(\frac{1}{2 \sqrt{2}}\) = 4

Now LHS is in the form a^{2} + 2ab Where b = \(\frac{1}{2 \sqrt{2}}\)

Adding b^{2} = \(\left(\frac{1}{2 \sqrt{2}}\right)^2\) on both sides we get

ii) 4x^{2} + 4\(\sqrt{3}\)x + 3= 0

Solution:

Given : 4x^{2} + 4\(\sqrt{3}\)x + 3 = 0

⇒ 4x^{2} + 4\(\sqrt{3}\)x = – 3

⇒ (2x)^{2} + 2(2x) \(\sqrt{3}\) = -3

LHS is of the form a^{2} + 2ab where b = \(\sqrt{3}\)

∴ Adding b^{2} = (\(\sqrt{3}\))^{2} = 3 on both sides, we get

⇒ (2x)^{2} + 2(2x)(\(\sqrt{3}\)) + (\(\sqrt{3}\))^{2}

= -3 + (\(\sqrt{3}\))^{2}

⇒ (2x + (\(\sqrt{3}\)))^{2} = -3 + 3 = 0

⇒ (2x + \(\sqrt{3}\))^{2} = 0

⇒ 2x + \(\sqrt{3}\) – 0

⇒ 2x + \(\sqrt{3}\) = 0

⇒ 2x = –\(\sqrt{3}\)

⇒ x = \(\frac{-\sqrt{3}}{2}\)

∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\)

iii) 5x^{2} – 7x – 6 = 0

Solution:

The given Q.E. is 5x^{2} – 7x – 6 = 0

⇒ 5x^{2} – 7x = 6

(\(\sqrt{5}\))^{2} – 2\(\sqrt{5}\). x.\(\left(\frac{7}{2 \sqrt{5}}\right)\) = 6

The LHS is of the form a^{2} – 2ab

where b = \(\frac{7}{2 \sqrt{5}}\)

Now adding b^{2} = \(\left(\frac{7}{2 \sqrt{5}}\right)^2\) on both sides we get

Note : If we take the Q.E. as 5x^{2} – 7x + 6 = 0, then we get the T.B. answer.

iv) x^{2} + 5 = -6x

Solution:

The given Q.E. is x^{2} + 5 = -6x

⇒ x^{2} + 6x = -5

⇒ (x)^{2} + 2. (x). 3 = -5

Now L.H.S. is of the form a^{2} + 2ab where b = 3.

Adding b^{2} = 3^{2} on both sides we get

x^{2} + 2(x)(3) + 3^{2} = -5 + 3^{2}

(x + 3)^{2} = -5 + 9 = 4

∴ x + 3 = \(\sqrt{4}\) = ±2

⇒ x = +2 – 3 or -2 -3

= -1 or -5 are the roots of the given Q.E.

Question 2.

Find the roots of the quadratic equations given in Q.l above by applying the quadratic formula,

i) 2x^{2} + x – 4 = 0

Solution:

Comparing this Q.E.with ax^{2} + bx + c = 0

a = 2, b = 1, c = -4

ii) 4x^{2} + 4\(\sqrt{3}\)x + 3 = 0

Solution:

Given : 4x^{2} + 4 \(\sqrt{3}\) x + 3 = 0

Here a = 4, b = \(\sqrt{3}\); c = 3

iii) 5x^{2} – 7x – 6 = 0

Solution:

Given : 5x^{2} – 7x – 6 = 0

Here a = 5, b = -7 and c = -6

iv) x^{2} + 5 = -6x

Solution:

Given : x^{2} + 5 = -6x

⇒ x^{2} + 6x + 5 = 0

Here a = 1; b = 6; c = 5

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

Question 3.

Find the roots of the following equations :

i) x – \(\frac{1}{x}\) = 3, x ≠ 0

Solution:

Given x – \(\frac{1}{x}\) = 3

⇒ \(\frac{x^2-1}{x}\) = 3

⇒ x^{2} – 1 = 3x

⇒ x^{2} – 3x – 1 = 0

Here a = 1; b = -3; c = 1

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

Solution:

⇒ x^{2} – 3x – 28 = -30

⇒ x^{2} – 3x – 28 + 30 = 0

⇒ x^{2} – 3x + 2 = 0

⇒ x^{2} – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2) (x – 1) = 0

⇒ x – 2 = 0 (or) x – 1 = O

⇒ x = 2 or x = 1

∴ x = 2 or 1.

Question 4.

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

Solution:

Let the present age of Rehman be x years.

3 years age Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x+5}\)

Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)

The sum of the reciprocals

⇒ x^{2} + 2x – 15 = 3(2x + 2)

⇒ x^{2} + 2x – 15 = 6x + 6

⇒ x^{2} + 2x – 15 – 6x – 6 = 0

⇒ x^{2} – 4x – 21 = 0

⇒ x^{2} – 7x + 3x – 21 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = -7 = 0 or x + 3 = 0

⇒ x = 7 or x = -3

But x can’t be negative

∴ x = 7

ie., present age of Rehman = 7 years.

Question 5.

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

Solution:

Sum of the marks in Mathematics and English = 30

Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x

If she got 2 more marks in Mathematics, then her marks would be x + 2.

It she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x

By problem (x + 2) (27 – x ) = 210

⇒ x(27 – x) + 2(27 – x) = 210

⇒ 27x – x^{2} + 54 – 2x = 210

⇒ -x^{2} + 25x + 54 = 210

⇒ x^{2} – 25x – 54 + 210 = 0

⇒ x^{2} – 25x + 156 = 0

⇒ x^{2} – 12x – 13x + 156 = 0

⇒ x(x – 12) – 13(x – 12) = 0

⇒ x – 12 = 0 or x – 13 = 0

⇒ x = 12 or x = 13

If x = 12, then marks in Mathematics =12

English = 30 – 12 = 18

If x = 13, then marks in Mathematics = 13

English = 30 – 13 = 17

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 the shorter side of the rectangular field = x m.

Then its longer side = x + 30 m

The diagonal of a rectangle is also the hypotenuse of the lower triangle.

Here the diagonal = x + 60

∴ By Pythagoras Theorem

(side)^{2} + (side)^{2} = (hypotenuse)^{2}

⇒ (x + 30)^{2} + x^{2} =(x + 60)^{2}

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

⇒ x^{2} – 60x – 2700 = 0

⇒ x^{2} – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30 (x – 90) = 0

⇒ (x – 90) (x + 30) = 0

⇒ x – 90 = 0(or) x + 30 = 0

⇒ x = 90 (or) x = -30

But ‘x’ can’t be negative.

∴ x = 90 m

i.e., the shorter side x = 90 m

Longer side x + 30 = 90 + 30 = 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 the large number be x.

8 times larger number = square of the small number = 8x

Square of the large number = x^{2}

By problem, x^{2} – 8x = 180

⇒ x^{2} – 8x – 180 = 0

⇒ x^{2} – 18x + 10x – 180 = 0

⇒ x(x – 18) + 10(x – 18) = 0

⇒ (x + 10) (x – 18) = 0

⇒ x + 10 = 0(or) x – 18 = 0

⇒ x = – 10(or) x = 18

If x = 18, then larger number = 18;

(small number)^{2} = 8 × (18) = 144

∴ Small number = \(\sqrt{144}\) = 12

The numbers are 18, 12

Note : Discard x = – 1o.

Question 8.

A train travels 360 km at a uniform speed. It the speed had been 5 km/br more, It would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

The distnce travelled = 360 km.

Let the speed of the train = x kmph.

Time taken to complete a journey =

By problem \(\frac{360}{x}\) – \(\frac{360}{x+5}\) = 1

⇒ 360\(\left(\frac{1}{x}-\frac{1}{x+5}\right)\) = 1

⇒ 360\(\left(\frac{x+5-x}{x(x+5)}\right)\)

⇒ \(\frac{5}{x^2+5 x}\) = \(\frac{1}{360}\)

⇒ x^{2} + 5x = 1800

⇒ x^{2} + 5x – 1800 = 0

⇒ x^{2} + 45x – 40x – 1800 = 0

⇒ x(x + 45) – 40(x + 45) = 0

⇒ (x + 45) (x – 40) = 0

⇒ x + 45 = 0 or x – 40 = 0

⇒ x = – 45 or x = 40

But x can’t be negative

∴ The speed of the train = 40 kmph.

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 the time taken by tap with smaller diameter alone to fill the tank = x hrs.

Then time taken by the tap with larger diameter = (x – 10) hrs.

By problem both taps worked for

9\(\frac{3}{8}\) = \(\frac{75}{8}\) hours

⇒ 75(x – 5) = 4(x^{2} – 10x)

⇒ 4x^{2} – 40x = 75x – 375

⇒ 4x^{2} – 115x + 375 = 0

⇒ 4x^{2} – 100x – 15x + 375 = 0

⇒ 4x(x – 25) – 15(x – 25) = 0

⇒ x – 25 = 0 (or) 4x – 15 = 0

⇒ x = 25 (or) x = \(\frac{15}{4}\)

x can’t be \(\frac{15}{4}\)

∴ x = 25

i.e., Time taken by the first tap = 25hrs

Time taken by the second tap

= 25 – 10 = 15 hrs.

Question 10.

An express train takes 1 hour less then 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/hr more than that of the passenger train, find the average speed of the two trains.

Solution:

Let the speed of the passenger train = x kmph

Then speed of the express train = x + 11 kmph.

Distance travelled = 132 km

⇒ x^{2} + 11x = 132 × 11

⇒ x^{2} + 11x – 1452 = 0

⇒ x^{2} + 44x – 33x – 1452 = 0

⇒ x(x + 44) – 33(x + 44) = 0

⇒ (x + 44) (x – 33) = 0

⇒ x + 44 = 0 (or) x – 33 = 0

⇒ x = -44 (or) x = 33

But x cant be negative.

∴ Speed of the passenger train

= x = 33 kmph.

Speed of the express train

= x + 11 = 44 kmph.

Question 11.

Sum of the areas of two squares is 468 m^{2}. If the difference of the perimeters is 24m, find the sides of the two square.

Solution:

Let the side of first square = x m say

Then perimeter of the first square

= 4x [∵ P = 4. side]

By Problem, perimeter of the second square = 4x + 24 (or) 4x – 24

∴ Side of the second square

⇒ \(\frac{4 x+24}{4}\) = \(\frac{4(x+6)}{4}\) = x + 6 (or)

\(\frac{4 x-24}{4}\) = x – 6

Now sum of the areas of the two squares is given as 468 m^{2}

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

x^{2} + x^{2} + 12x + 36 = 468

2x^{2} + 12x + 36 – 468 = 0

2x^{2} + 12x – 432 = 0

x^{2} + 6x – 216 = 0

x^{2} + 18x – 12x – 216 = 0

x(x + 18) – 12(x + 18) = 0

(x + 18) (x – 12) = 0

⇒ x + 18 = 0 (or) x – 12 = 0

⇒ x = – 18 (or) 12

But x can’t be negative

∴ x = 12

i.e., side of the first square = 12

∴ Perimeter = 4 × 12 = 48

∴ Perimeter of the second square = 48 + 24 = 72

∴ Side of the second square = \(\frac{72}{4}\) = 18 m.

(or)

x^{2} + (x – 6)^{2} = 468

x^{2} + x^{2} – 12x + 36 = 468

2x^{2} – 12x – 432 = 0

x^{2} – 6x – 216 = 0

x^{2}– 18x + 12x – 216 = 0

x(x – 18) + 12(x – 18) = 0

(x – 18) (x + 12) = 0

⇒ x – 18 = 0 (or) x + 12 = 0

⇒ x = 18 (or) -12

But x can’t be negative

∴ x = 18

i.e., side of the first square = 18 m

Perimeter = 4 × 18 = 72

Perimeter of the second square

= 72 – 24 = 48

∴ Side of the second square = \(\frac{48}{4}\) = 12 m

i. e., In any way, the sides of the squares are 12m, 18m.

Question 12.

S = ut – \(\frac{1}{2}\)gt^{2} is a formula which gives the distance S in meters travelled by a ball from the thrower’s hands if it is thrown upwards with an initial velocity of u m/s after a time of t seconds, g is the acceleration due to gravity and is 9.8 m/s^{2}

(i) If a ball is thrown upwards at 14 m/s, how high has it gone after 1 second ?

(ii) How long does it take for the ball to reach a height of 5 meters ?

(iii) Why are there two possible times to reach a height of 5 meters ?

Solution:

S = ut – \(\frac{1}{2}\)gt^{2}

i) u = 14 m/s; t = 1 sec; g = 9.8 m/s^{2}

S = 14 × 1 – \(\frac{1}{2}\) × 9.8 × (1)^{2}

= 14 – \(\frac{1}{2}\) × 9.8 = 14 – 4.9 = 9.1m.

ii) S = 5 m.

5 = 14t – 4.9 t^{2}

4.9 t^{2} – 14t + 5 = 0

\(\frac{49}{10}\)t^{2} – 14t + 5 = 0

49t^{2} – 140t + 50 = 0

a = 49, b = -140, c = 50

iii) When the body goes upwards t value is 2.4 sec.

When the body comes downwards t value is 0.4 sec.

∴ So there are two possible times to reach a height of 5 meters.

Question 13.

If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals ? Is there a polygon with 50 diagonals ?

Solution:

Given : Number of diagonals of a polygon with n-sides = \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\)

No. of diagonals of a given polygon = 65

i.e., \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 65

where n is number of sides of the polygon

⇒ n^{2} – 3n = 2 × 65

⇒ n^{2} – 3n – 130 = 0

⇒ n^{2} – 13n + 10n – 130 = 0

⇒ n(n – 13) + 10(n – 13) = 0

⇒ (n – 13) (n + 10) = 0

⇒ (n – 13) = 0 (or) n + 10 = 0

⇒ n = 13 (or) n = -10

But n can’t be negative.

∴ n = 13 (i.e.) number of sides = 13.

Also to check 50 as the number of diagonals of a polygon

∴ \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 50

⇒ n^{2} – 3n = 100

⇒ n^{2} – 3n – 100 = 0

There is no real value of n for which the above equation is satisfied.

∴ There can’t be a polygon with 50 diagonals.