Important Questions Class 10 Maths Chapter 4 Quadratic Equations

Important questions with solutions of chapter 4 quadratic equations for class 10 Maths are available here. Students who are preparing for CBSE board exam for the session 2019-2020, can practice these questions to score good marks. These questions have been formulated as per NCERT book, so there will no problems here which are out of the syllabus. Students can also access here important questions for all the chapters of Class 10 Maths.

The questions here are based on quadratic equations, how to solve quadratic equations and finding the roots by factorisation methods. Solving these questions will help students to do better revision for the exams.

Important Questions with Solutions for Quadratic Equations

Let us solve here some of the questions which are important with respect to class 10th Maths board exam.

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.

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 to cover the same distance.
Therefore, time taken to travel 480 km = 480x + 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 – 8x – 1280 = 0

Hence, 3x2 – 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 (x – 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 =110 or x =110

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

Solutions: Let us say, the two consecutive positive integers be x and x + 1.
Therefore, as per the given questions,
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
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.

Solutions: Let us say, the shorter side of the rectangle is 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 + 30 m
Therefore,
√[x2+(x+30)2]=x+60

Squaring on both the sides, we get;
⇒ 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: 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.

Solutions: Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x2 and y2 respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6

Also, x2 + y2 = 468

⇒ (6 + y2) + y2 = 468

⇒ 36 + y2 + 12y + y2 = 468

⇒ 2y2 + 12y + 432 = 0

⇒ y2 + 6y – 216 = 0

⇒ y2 + 18y – 12y – 216 = 0

⇒ y(y +18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18

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

Solutions:

(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 400m2? 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 3×2– 2x +1/3= 0 and hence find the nature of its roots. Find them, if they are real.
Solution : 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

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  1. THANKS ,IT IS VERY USEFUL FOR ME

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