**1. Check whether the following are quadratic equations: **

Solution:

(i) (x+1)^{2} = 2(x – 3)

x^{2} + 2x + 1 = 2x – 6

x^{2} + 7 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is a quadratic equation.

(ii) x^{2} – 2x = (-2)(3 – x)

x^{2} – 2x = -6 + 2x

x^{2} – 4x + 6 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is a quadratic equation.

* *

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

x^{2} + x – 2x + 2 = x^{2} + 3x – x – 3

x^{2} – x – 2 = x^{2} + 2x – 3

-3x + 1 = 0

3x – 1 = 0

It is not of the form ax^{2 }+ bx + c.

Hence, the given equation is not a quadratic equation.

* *

(iv) (x – 3)(2x + 1) = x (x + 5)

2x^{2} + x – 6x – 3 = x^{2} + 5x

x^{2} – 5x – 5x – 3 = 0

x^{2} – 10x – 3 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is a quadratic equation.

* *

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

2x^{2} – 6x – x + 3 = x^{2} – x + 5x – 5

x^{2} – 7x + 3 = 4x – 5

x^{2} – 11x + 8 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is a quadratic equation

(vi) x^{2} + 3x + 1 = (x – 2)^{2}

x^{2} – 3x + 1 = x^{2} – 4x + 4

7x – 3 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is not a quadratic equation.

(vii) (x + 2)^{3} = 2x(x^{2 }– 1)

x^{3} + 8 + 6x^{2} + 12x = 2x^{3} – 2x

x^{3} – 14x – 6x^{2} – 8 = 0

It is not of the form ax^{2 }+ bx + c.

Hence, the given equation is not a quadratic equation.

(vii) x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

x^{3} – 4x^{2} – x + 1 = x^{3} – 8 – 6x^{2} + 12x

2x^{2} – 13x + 9 = 0

It is of the form ax^{2 }+ bx + c.

Hence, the given equation is a quadratic

**2. 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) The product of two consecutive positive integers is 306. We need to find the integers. **

**(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. **

**(iv) 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. We need to find the speed of the train.**

Solution:

(i) Let the breadth of the plot be ‘x’ m.

Hence, the length of the plot is (2x + 1) m.

Area of a rectangle = Length × Breadth

∴ 528 = x (2x + 1)

2x² + x – 528 =0

(ii) Let the consecutive integers be ‘x’ and ‘x + 1’.

It is given that their product is 306.

x(x + 1) = 306

2x² + x – 306 = 0

(iii)Let Rohan’s age be ‘x’.

Hence, his mother’s age = x + 26 3 years

hence, Rohan’s age = x + 3

Mother’s age = x + 26 + 3 = x + 29

It is given that the product of their ages after 3 years is 360.

(x + 3)(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x – 273 = 0

(iv) Let the speed of train be x km/h.

Time taken to travel 480 km = 480/x hrs

In second condition, let the speed of 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

## 5 responses to “Quadratic Equations – Exercise 4.1 – Class 10”

(…I am so glad I am out of school, these past 45 years!)

In spite of the above thought, I remember now why I found university maths so fascinating! You’ve a very helpful blog for those still involved in the study of math.

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Pearl

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And I thank you as well😊

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Haha!

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