CLASS 10 IMPORTANT QUESTIONS CHAPTER 4 (QUADRATIC EQUATIONS) NCERT

 CLASS 10 IMPORTANT QUESTIONS CHAPTER 4 (QUADRATIC EQUATIONS) NCERT 

Q.1: 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?

Solution:

(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.

Q.2: Find the roots of quadratic equations by factorisation:

(i) √2 x² + 7x + 5√2=0

(ii) 100x² – 20x + 1 = 0

Solution:

(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

Q.3: Find two consecutive positive integers, the sum of whose squares is 365.

Solution:

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.

Q.4: 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

Solution:

(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.

Q.5: 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 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

Q.6 : Solve the quadratic equation 2x² – 7x + 3 = 0 by using quadratic formula.

Solution: 

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

Q.7: 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.

Solution:

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.

Q.8: 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

Solution:

(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.

Q.9: Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.

Solution:

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.

Q.10: 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.
Solution:

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