Quadratic Equations – Exercise 9.1
- 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) x2 – 29/4 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
- 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) x2 – 29/4 x + 5 = 0
Solution:
Given equation is x2 –29/4 x + 5 = 0
The equation is of the form ax2 + bx + c = 0.
Therefore, x2 –29/4 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
