Class 10 Maths Chapter 4 Practice Question Answers - Quadratic Equations

Q1. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. What is the speed of the train?

Sol:
(i) Let us consider,

The breadth of the rectangular plot is x m.

Thus, the length of the plot = (2x + 1) m

As we know,

Area of rectangle = length × breadth = 528 m²

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x² + x = 528

⇒ 2x² + x – 528 = 0

Hence, 2x² + x – 528 = 0, is the required equation which represents the given situation.

(ii) Let us consider,

speed of train = x km/h

And

Time taken to travel 480 km = 480 (x) km/h

As per second situation, the speed of train = (x – 8) km/h

As given, the train will take 3 hours more to cover the same distance.

Therefore, time taken to travel 480 km = (480/x) + 3 km/h

As we know,

Speed × Time = Distance

Therefore,

(x – 8)[(480/x) + 3] = 480

⇒ 480 + 3x – (3840/x) – 24 = 480

⇒ 3x – (3840/x) = 24

⇒ 3x² – 24x – 3840 = 0

⇒ x² – 8x – 1280 = 0

Hence, x² – 8x – 1280 = 0 is the required representation of the problem mathematically.

Q2. Find the roots of quadratic equations by factorisation:
(i) √2 x2 + 7x + 5√2=0
(ii) 100x² – 20x + 1 = 0

Sol: 

(i)  √2 x² + 7x + 5√2=0

Considering the L.H.S. first,

⇒ √2 x² + 5x + 2x + 5√2

⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (√2x + 5)(x + √2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

⇒ x = -5/√2 or x = -√2

(ii) Given, 100x² – 20x + 1=0

Considering the L.H.S. first,

⇒ 100x² – 10x – 10x + 1

⇒ 10x(10x – 1) -1(10x – 1)

⇒ (10x – 1)²

The roots of this equation, 100x² – 20x + 1=0, are the values of x for which (10x – 1)²= 0

Therefore,

(10x – 1) = 0

or (10x – 1) = 0

⇒ x =1/10 or x =1/10

Q3. Find two consecutive positive integers, the sum of whose squares is 365.

Sol: 

Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given statement,

x² + (x + 1)2 = 365

⇒ x² + x2 + 1 + 2x = 365

⇒ 2x² + 2x – 364 = 0

⇒ x² + x – 182 = 0

⇒ x² + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

So, x + 1 = 13 + 1 = 14

Therefore, the two consecutive positive integers will be 13 and 14.

Q4. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x +3 = 0
(ii) 2x² + x – 4 = 0
Sol:

(i) 2x² – 7x + 3 = 0

⇒ 2x² – 7x = – 3

Dividing by 2 on both sides, we get

⇒x²–7x/2 = -3/2

⇒x² – 2 × x × 7/4 = -3/2

On adding (7/4)2 to both sides of above equation, we get

⇒ (x)² – 2 × x × 7/4 + (7/4)² = (7/4)² – (3/2)

⇒ (x – 7/4)² = (49/16) -(3/2)

⇒ (x – 7/4)² = 25/16

⇒ (x – 7/4) = ± 5/4

⇒ x = 7/4 ±5/4

⇒ x = 7/4 +5/4 or x = 7/4-5/4

⇒ x =12/4 or x =2/4

⇒ x = 3 or 1/2

(ii) 2x² + x – 4 = 0

⇒ 2x2 + x = 4

Dividing both sides of the above equation by 2, we get

⇒ x² + x/2 = 2

⇒ (x)² + 2 × x × 1/4 = 2

Now on adding (1/4)2 to both sides of the equation, we get,

⇒ (x)² + 2 × x × 1/4 + (1/4)² = 2 + (1/4)²

⇒ (x + 1/4)² = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = ± √33 – 1/4

Therefore, either x = √33-1/4 or x = -√33-1/4.

Q5. 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.

Sol: 

Let us say, the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of the rectangle = √[x²+(x+30)²] As given, the length of the diagonal is = x + 60 m

⇒ x² + (x + 30)2 = (x + 60)²

⇒ x² + x2 + 900 + 60x = x² + 3600 + 120x

⇒ x² – 60x – 2700 = 0

⇒ x² – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x -90)

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

Q6. Solve the quadratic equation 2x² – 7x + 3 = 0 by using quadratic formula.

Sol:  

2x² – 7x + 3 = 0

On comparing the given equation with ax² + bx + c = 0, we get,

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

By using quadratic formula, we get,

x = [-b±√(b² – 4ac)]/2a

⇒ x = [7±√(49 – 24)]/4

⇒ x = [7±√25]/4

⇒ x = [7±5]/4

Therefore,

⇒ x = 7+5/4 or x = 7-5/4

⇒ x = 12/4 or 2/4

∴ x = 3 or ½

Q7. The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol:

Sum of the areas of two squares is 468 m².

∵ x² + y² = 468 . ………..(1) [ ∵ area of square = side²]

→ The difference of their perimeters is 24 m.

∵ 4x – 4y = 24 [ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24

⇒ x – y = 24/4

⇒ x – y = 6

∴ y = x – 6 ……….(2)

From equation (1) and (2),

∵ x² + ( x – 6 )² = 468

⇒ x² + x² – 12x + 36 = 468

⇒ 2x² – 12x + 36 – 468 = 0

⇒ 2x² – 12x – 432 = 0

⇒ 2( x² – 6x – 216 ) = 0

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0

⇒ x( x – 18 ) + 12( x – 18 ) = 0

⇒ ( x + 12 ) ( x – 18 ) = 0

⇒ x + 12 = 0 and x – 18 = 0

⇒ x = – 12m [ rejected ] and x = 18m

∴ x = 18 m

Put the value of ‘x’ in equation (2),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12 m

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

Q8. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0

(ii) kx (x – 2) + 6 = 0

Sol:

(i) 2x² + kx + 3 = 0

Comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = k and c = 3

As we know, Discriminant = b² – 4ac

= (k)² – 4(2) (3)

= k² – 24

For equal roots, we know,

Discriminant = 0

k² – 24 = 0

k² = 24

k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx² – 2kx + 6 = 0

Comparing the given equation with ax² + bx + c = 0, we get

a = k, b = – 2k and c = 6

We know, Discriminant = b² – 4ac

= ( – 2k)² – 4 (k) (6)

= 4k² – 24k

For equal roots, we know,

b² – 4ac = 0

4k² – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x²‘ and ‘x‘.

Therefore, if this equation has two equal roots, k should be 6 only.

Q9. Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.

Sol:

Let the length and breadth of the park be L and B.

Perimeter of the rectangular park = 2 (L + B) = 80

So, L + B = 40

Or, B = 40 – L

Area of the rectangular park = L × B = L(40 – L) = 40L – L² = 400

L² – 40 L + 400 = 0,

which is a quadratic equation.

Comparing the equation with ax² + bx + c = 0, we get

a = 1, b = -40, c = 400

Since, Discriminant = b² – 4ac

=>(-40)² – 4 × 400

=> 1600 – 1600

= 0

Thus, b² – 4ac = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

Root of the equation,

L = –b/2a

L = (40)/2(1) = 40/2 = 20

Therefore, length of rectangular park, L = 20 m

And breadth of the park, B = 40 – L = 40 – 20 = 20 m.

Q10. Find the discriminant of the equation 3x²– 2x +1/3= 0 and hence find the nature of its roots. Find them, if they are real.

Sol:

Given,

3x²– 2x +1/3= 0

Here, a = 3, b = – 2 and c = 1/3

Since, Discriminant = b2 – 4ac

= (– 2)² – 4 × 3 × 1/3

= 4 – 4 = 0.

Hence, the given quadratic equation has two equal real roots.

The roots are -b/2a and -b/2a.

2/6 and 2/6

or

1/3, 1/3

Q11. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.

Sol:

Let the duration of the flight be x hours.

According to the given,

(600/x) – [600/(x + 1/2) = 200

(600/x) – [1200/(2x + 1)] = 200

[600(2x + 1) – 1200x]/ [x(2x + 1)] = 200

(1200x + 600 – 1200x)/ [x(2x + 1)] = 200

600 = 200x(2x + 1)

x(2x + 1) = 3

2x² + x – 3 = 0

2x² + 3x – 2x – 3 = 0

x(2x + 3) – 1(2x + 3) = 0

(2x + 3)(x – 1) = 0

2x + 3 = 0, x – 1 = 0

x = -3/2, x = 1

Time cannot be negative.

Therefore, x = 1

Hence, the original duration of the flight is 1 hr.

Q12. If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.

Sol:

Given that x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0.

⇒ (3)2 – 2k(3) – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ 3 – 6k = 0

⇒ 6k = 3

⇒ k = 1/2

Therefore, the value of k is ½.

Q.13: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.

Sol:

Given quadratic equation is:

px2 – 14x + 8 = 0

Let α and 6α be the roots of the given quadratic equation.

Sum of the roots = -coefficient of x/coefficient of x2

α + 6α = -(-14)/p

7α = 14/p

α  = 2/p….(i)

Product of roots = constant term/coefficient of x2

(α)(6α) = 8/p

6α² = 8/p

Substituting α = 2/p from (i),

6 × (2/p)² = 8/p

24/p² = 8/p

3/p = 1

p = 3

Therefore, the value of p is 3.

Q14. Solve for x: [1/(x + 1)] + [3/(5x + 1)] = 5/(x + 4); x ≠ -1, -⅕, -4

Sol:

Given,

[1/(x + 1)] + [3/(5x + 1)] = 5/(x + 4); x ≠ -1, -⅕, -4

Let us take the LCM of denominators and cross multiply the terms.

[1(5x + 1) + 3(x + 1)]/ [(x + 1)(5x + 1)] = 5/(x + 4)

[5x + 1 + 3x + 3]/ [5x² + x + 5x + 1] = 5/(x + 4)

(8x + 4)(x + 4) = 5(5x² + 6x + 1)

8x² + 32x + 4x + 16 = 25x² + 30x + 5

25x² + 30x + 5 – 8x2 – 36x – 16 = 0

17x² – 6x – 11 = 0

17x² – 17x + 11x – 11 = 0

17x(x – 1) + 11(x – 1) = 0

(17x + 11)(x – 1) = 0

17x + 11 = 0, x – 1 = 0

x = -11/17, x = 1

Q.15: If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Sol:

Given that -5 is a root of the quadratic equation 2x² + px – 15 = 0.

⇒ 2(-5)² + p(-5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 35 – 5p = 0

⇒ 5p = 35

⇒ p = 7

Also, the quadratic equation p(x² + x) + k = 0 has equal roots.

Substituting p = 7 in p(x² + x) + k = 0,

7(x² + x) + k = 0

7x² + 7x + k = 0

Comparing with the standard form ax² + bx + c = 0,

a = 7, b = 7, c = k

For equal roots, discriminant is equal to 0.

b² – 4ac = 0

(7)² – 4(7)(k) = 0

49 – 28k = 0

28k = 49

k = 7/4

Therefore, the value of k is 7/4.