NCERT Exemplar: Quadratic Equations - 2 | Mathematics (Maths) Class 10 PDF Download

Exercise 4.4

Q.1. Find whether the following equations have real roots. If real roots exist, find them.

(i) 8x2 + 2x - 3 = 0
(ii) -2x2 + 3x + 2 = 0

(iii) 5x2 - 2x - 10 = 0
(iv)

(v) x2 + 5√5x- 70 = 0

(i)

a quadratic equation ax² + bx + c = 0 has two distinct real roots
if its discriminant D = b² - 4ac > 0, 
one real root (repeated) if D = 0, 
 no real roots if D < 0.

a = 8, b = 2, c = -3

D = b² - 4ac = 2² - 4·8·(-3)

D = 4 + 96 = 100 > 0

Therefore two distinct real roots exist.

x = [-b ± √D]/(2a)

x = (-2 ± 10)/16

x = (-2 + 10)/16 = 8/16 = 1/2

x = (-2 - 10)/16 = -12/16 = -3/4

Hence the roots are 1/2 and -3/4.

(ii)

Given -2x² + 3x + 2 = 0. Multiply by -1 to write in standard form:

2x² - 3x - 2 = 0

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

D = (-3)² - 4·2·(-2) = 9 + 16 = 25 > 0

Therefore two distinct real roots exist.

x = [-b ± √D]/(2a)

x = (3 ± 5)/4

x = (3 + 5)/4 = 8/4 = 2

x = (3 - 5)/4 = -2/4 = -1/2

Hence the roots are 2 and -1/2.

(iii)

Given 5x² - 2x - 10 = 0

a = 5, b = -2, c = -10

D = (-2)² - 4·5·(-10) = 4 + 200 = 204 > 0

Therefore two distinct real roots exist.

x = [-b ± √D]/(2a)

x = (2 ± √204)/10

√204 = 2√51, so x = (2 ± 2√51)/10 = (1 ± √51)/5

Hence the roots are (1 + √51)/5 and (1 - √51)/5.

(iv)

Given 1/(2x - 3) + 1/(x - 5) = 1. Domain: x ≠ 3/2, x ≠ 5.

Find common denominator and combine:

(x - 5) + (2x - 3) = (2x - 3)(x - 5)

3x - 8 = 2x² - 13x + 15

Bring all terms to one side:

0 = 2x² - 13x + 15 - 3x + 8

0 = 2x² - 16x + 23

a = 2, b = -16, c = 23

D = (-16)² - 4·2·23 = 256 - 184 = 72 > 0

Therefore two distinct real roots exist.

x = [-b ± √D]/(2a)

x = (16 ± √72)/4

√72 = 6√2, so x = (16 ± 6√2)/4 = (8 ± 3√2)/2

Hence the roots are (8 + 3√2)/2 and (8 - 3√2)/2, provided they do not equal 3/2 or 5 (they do not).

(v)

Given x² + 5√5 x - 70 = 0

a = 1, b = 5√5, c = -70

D = (5√5)² - 4·1·(-70) = 125 + 280 = 405 > 0

Therefore two distinct real roots exist.

x = [-b ± √D]/(2a)

√405 = 9√5, so x = (-5√5 ± 9√5)/2

x = (4√5)/2 = 2√5 and x = (-14√5)/2 = -7√5

Hence the roots are 2√5 and -7√5.

Q.2. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Let the natural number be x.

According to the question:

x² - 84 = 3(x + 8)

x² - 84 = 3x + 24

x² - 3x - 108 = 0

Factorise:
x² – 12x + 9x – 108 = 0
x(x – 12) + 9(x – 12) = 0

(x - 12)(x + 9) = 0

x = 12 or x = -9

Natural number must be positive, so x = 12.

Let the natural number be x.
When the number increased by 12 = x + 12
Reciprocal of the number = 1/x
According to the question, we have,

x + 12 = 160/x

x(x + 12) = 160

x² + 12x - 160 = 0

Factorise:
x² + 20x – 8x – 160 = 0
x( x + 20) – 8( x + 20)=  0

(x + 20)(x - 8) = 0

x = -20 or x = 8

Natural number is positive, so x = 8.

Q.4. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Let the original speed be x km/h.

Time = distance/speed.

Time at speed x: 360/x hours.

Time at speed x + 5: 360/(x + 5) hours.

Given difference 48 minutes = 48/60 = 4/5 hour:

360/x - 360/(x + 5) = 4/5

Multiply both sides by x(x + 5): 360(x + 5) - 360x = (4/5) x(x + 5)

Left side simplifies: 360x + 1800 - 360x = 1800

So 1800 = (4/5) x(x + 5)

Multiply both sides by 5/4: x(x + 5) = 1800·5/4 = 2250

x² + 5x - 2250 = 0

Discriminant D = 25 + 9000 = 9025 = 95²

x = [-5 ± 95]/2

x = (90)/2 = 45 or x = (-100)/2 = -50

Speed cannot be negative, so original speed x = 45 km/h.

Q.5. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Let Zeba's present age be x.
According to the question,

(x - 5)² = 11 + 5x

x² - 10x + 25 = 5x + 11

x² - 15x + 14 = 0

Factorise:
x² - 14x - x + 14 = 0
x(x -14) -1(x - 14) = 0

(x - 1)(x - 14) = 0

x = 1 or x = 14

If x = 1 then x - 5 = -4 (not meaningful for "younger by 5 years" in ordinary age context), so take x = 14 years.

Q.6. At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Let Nisha's present age be x years.

Then Asha's present age is x² + 2 years.

Years for Nisha to grow to Asha's present age = (x² + 2) - x = x² - x + 2.

Asha's age when Nisha attains Asha's present age = (x² + 2) + (x² - x + 2) = 2x² - x + 4.

Given this equals one year less than ten times present age of Nisha:

2x² - x + 4 = 10x - 1

2x² - 11x + 5 = 0

Factorise:
2x² + 10x - x + 5 = 0
2x(x + 5) - 1(x - 5) = 0

(2x - 1)(x - 5) = 0

x = 1/2 or x = 5

Nisha's age must be a reasonable whole number, so x = 5 years.

Asha's present age = x² + 2 = 25 + 2 = 27 years.

Given, the dimensions of the rectangular lawn is 50m × 40m.

Area of the grass surrounding the pond is 1184 m².

We have to determine the length and breadth of the pond.

Let ABCD be the rectangular lawn and EFGH be the rectangular pond.
Let x m be the width of the grass area around the pond.
So, length of the lawn = 50 m
Breadth of the lawn = 40 m
Length of the pond = (50 - 2x)m
Breadth of the pond = (40 - 2x)m
Area of grass = area of lawn - area of pond
1184 = (50 × 40) - [(50 - 2x) × (40 - 2x)]
1184 = 2000 - [2000 - 100x - 80x + 4x²]
1184 = 2000 - 2000 + 100x + 80x - 4x²
1184 = 180x - 4x²
So, 4x² - 180x + 1184 = 0
Dividing by 4,
x² - 45x + 296 = 0
On factoring,
x² - 37x - 8x + 296 = 0
x(x - 37) - 8(x - 37) = 0
(x - 8)(x - 37) = 0
Now, x - 8 = 0
x = 8
Also, x - 37 = 0
x = 37
When x = 37, the length of the pond = 50 - 2(37) = 50 - 74 = -24
So, x = 37 is not possible.
Therefore, the width of the grass area is 8 m.
Now, length of the pond = 50 - 2(8) = 50 - 16 = 34 m
Breadth of the pond = 40 - 2(8) = 40 - 16 = 24m
Therefore, the length and breadth of the pond is 34 and 24m.

At t minutes past 2 pm the remaining time to 3 pm is 60 - t minutes.

Given 60 - t = (t²/4) - 3.

Multiply both sides by 4:

240 - 4t = t² - 12.

Bring all terms to one side:

t² + 4t - 252 = 0.

Factorise:

(t + 18)(t - 14) = 0.

t = -18 or t = 14.

Negative time is not acceptable here, so t = 14 minutes.