Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.1

  1. Check whether the following are quadratic equations:

(i) x2 – x = 0

(ii) x2 = 8

(iii) x2 + 1/2 x = 0

(iv) 3x – 10 = 0

(v) x229/4 x + 5 = 0

(vi) 5 – 6x = 2/5x2

(vii) √2x2 + 3x = 0

(viii) √3x = 22/13

(ix)x3 – 10x + 74 = 0

(x) x2 – y2 = 0

2.Simplify the following equations and check whether they are quadratic equations:

(i) x(x + 6) = 0

(ii) (x – 4)(2x – 3) = 0

(iii) (x + 2)(x – 7) = 0

(iv) (x + 1)2 = 2(x – 3)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) (x + 2)3 = 2x(x2 – 1)

3.Represent the following in the form of quadratic equations:

(i) The product of two consecutive integers is 306

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.


Quadratic Equations – Exercise 9.1

  1. Check whether the following are quadratic equations:

(i) x2 – x = 0

Solution:

Given equation is x2 – x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x2 – x = 0 is quadratic equation.


 (ii) x2 = 8

Solution:

Given equation is x2 = 8

⇒ x2 + 0.x – 8 = 0

The equation is of the form ax2 + bx + c = 0, where coefficient of x is 0

Therefore, x2 – 8 = 0 is quadratic equation.


 (iii) x2 + 1/2 x = 0

Solution:

Given equation is x2 +1/2 x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x2 +1/2 x = 0 is quadratic equation.


 (iv) 3x – 10 = 0

Solution:

Given equation is 3x – 10 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore,  3x – 10 = 0 is not a quadratic equation.


 (v) x229/4 x + 5 = 0

Solution:

Given equation is x229/4 x + 5 = 0

The equation is of the form ax2 + bx + c = 0.

Therefore, x229/4 x + 5 = 0 is quadratic equation.


 (vi) 5 – 6x = 2/5x2

Solution:

Given equation is 5  – 6x = 2/5x2

2/5x2 + 6x – 5 = 0

The equation is of the form ax2 + bx + c = 0.

Therefore, 5  – 6x = 2/5x2 is quadratic equation.


 (vii) √2x2 + 3x = 0

Solution:

Given equation is √2x2 + 3x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, √2x2 + 3x = 0 is quadratic equation.


 (viii) √3x = 22/13

Solution:

Given equation is √3x = 22/13

⇒ √3x – 22/13 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore, √3x = 22/13 is not a quadratic equation.


 (ix)x3 – 10x + 74 = 0

Solution:

Given equation is x3 – 10x + 74 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore, x3 – 10x + 74 = 0 is not a quadratic equation.


 (x) x2 – y2 = 0

Solution:

Given equation is x2 – y2 = 0

The equation is of the form ax2 + bx + c = 0, where coefficient of x = 0

Therefore, x2 – y2 = 0 is a quadratic equation.


2.Simplify the following equations and check whether they are quadratic equations:

(i) x(x + 6) = 0

Solution:

x(x + 6) = 0

x2 + 6x = 0

The given equation is x2 + 6x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x(x + 6) = 0 is a quadratic equation.


 (ii) (x – 4)(2x – 3) = 0

Solution:

(x – 4)(2x – 3) = 0

2x2 – 3x – 8x + 12 = 0

2x2 – 11x + 12 = 0

Therefore, the given equation is (x – 4)(2x – 3) = 0

The equation is of the form ax2 + bx + c = 0

Therefore, (x – 4)(2x – 3) = 0 is a quadratic equation.


 (iii) (x + 2)(x – 7) = 0

Solution:

(x + 2)(x + 7) = 0

x2 + 7x + 2x + 14 = 0

x2 + 9x + 14 = 0

Therefore, the given equation is (x + 2)(x + 7) = 0

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 2)(x + 7) = 0 is a quadratic equation.


 (iv) (x + 1)2 = 2(x – 3)

Solution:

(x + 1)2 = 2(x – 3)

x2 + 2x + 1 = 2x – 6

x2 + 2x  – 2x + 1 – 6 = 0

x2  – 5 = 0

Therefore, the given equation is (x + 1)2 = 2(x – 3)

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 1)2 = 2(x – 3) is a quadratic equation.


 (v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution:

(2x – 1)(x – 3) = (x + 5)(x – 1)

2x2 – 6x – x + 3 = x2 – x + 5x – 5

2x2 – 6x – x + 3 – x2 + x – 5x + 5 = 0

x2 – 12x + 8 = 0

Therefore, the given equation is (2x – 1)(x – 3) = (x + 5)(x – 1)

The equation is of the form ax2 + bx + c = 0

Therefore, (2x – 1)(x – 3) = (x + 5)(x – 1) is a quadratic equation.


 (vi) (x + 2)3 = 2x(x2 – 1)

Solution:

(x + 2)3 = 2x(x2 – 1)

(x + 2)(x2 + 2x + 4) = 2x3 – 2x

x3 + 2x2 + 4x + 2x2 + 4x + 8 – 2x3 – 2x = 0

-x3 + 4x2 + 6x + 8 = 0

Therefore, the given equation is (x + 2)3 = 2x(x2 – 1)

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 2)3 = 2x(x2 – 1) is a quadratic equation.


3.Represent the following in the form of quadratic equations:

(i) The product of two consecutive integers is 306

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

Solution:

 

 

(i) The product of two consecutive integers is 306

Let two consecutive numbers be x and x+1.

Then, x(x + 1) = 306

x2 + x – 306 = 0

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

Area of rectangle = length x breadth

If length of the rectangular park = x m

then, breadth of the rectangular park = (2x + 1) m

(2x + 1)x = 528

2x2 + x – 528 = 0

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

Let the speed of train be x km/h

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

In second condition let the speed of the train = (x – 8) km/h

It is also given that the train will take 3 hours to cover the same distance.

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

Speed x time = distance

(x – 8)(480/x + 3) = 480

480 + 3x – 3840/x – 24 = 0

3x – 3840/x – 24 = 0

3x2 – 3840 – 24x = 0

x2 – 8x – 1280 = 0


Next Exercise – Quadratic equations – Exercise 9.2 – Class X