**NCERT Solutions for Class 10 Maths Chapter 4**

**Quadratic Equations Exercise 4.1**

**Page 73**

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

**(i) (x + 1) ^{2} = 2(x – 3)**

Given,

(x + 1)^{2} = 2(x – 3)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

⇒ x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 7 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

∴ the given equation is quadratic equation.

**(ii) x ^{2} – 2x = (–2) (3 – x)**

Given, x^{2} – 2x = (–2) (3 – x)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

⇒ x^{2 }–^{ }2x = -6 + 2x

⇒ x^{2 }– 4x + 6 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

∴ the given equation is quadratic equation.

**(iii) (x – 2)(x + 1) = (x – 1)(x + 3)**

Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

⇒ x^{2 }– x – 2 = x^{2 }+ 2x – 3

⇒ 3x – 1 = 0

Since the above equation is not in the form of ax^{2} + bx + c = 0.

∴ the given equation is not a quadratic equation.

**(iv) (x – 3)(2x +1) = x(x + 5)**

Given, (x – 3)(2x +1) = x(x + 5)

By using the formula for (a+b)^{2}=a^{2}+2ab+b^{2}

⇒ 2x^{2 }– 5x – 3 = x^{2 }+ 5x

⇒ x^{2 }– 10x – 3 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

∴ the given equation is quadratic equation.

**(v) (2x – 1)(x – 3) = (x + 5)(x – 1)**

Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

By using the formula for (a+b)^{2}=a^{2}+2ab+b^{2}

⇒ 2x^{2 }– 7x + 3 = x^{2 }+ 4x – 5

⇒ x^{2 }– 11x + 8 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

∴ the given equation is quadratic equation.

**(vi) x ^{2} + 3x + 1 = (x – 2)^{2}**

Given, x^{2} + 3x + 1 = (x – 2)^{2}

By using the formula for (a+b)^{2}=a^{2}+2ab+b^{2}

⇒ x^{2} + 3x + 1 = x^{2} + 4 – 4x

⇒ 7x – 3 = 0

Since the above equation is not in the form of ax^{2} + bx + c = 0.

∴ the given equation is not a quadratic equation.

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

Given, (x + 2)^{3} = 2x(x^{2} – 1)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

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

⇒ x^{3} + 14x – 6x^{2} – 8 = 0

Since the above equation is not in the form of ax^{2} + bx + c = 0.

∴ the given equation is not a quadratic equation.

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

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

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

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

⇒ 2x^{2} – 13x + 9 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

∴ the given equation is quadratic equation.

**2. Represent the following situations in the form of quadratic equations:**

**(i) The area of a rectangular plot is 528 m ^{2}. 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.**

Let us consider,

Breadth of the rectangular plot = x m

Thus, the length of the plot = (2x + 1) m.

As we know,

Area of rectangle = length × breadth = 528 m^{2}

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x^{2} + x =528

⇒ 2x^{2} + x – 528 = 0

∴ the length and breadth of plot, satisfies the quadratic equation, 2x^{2} + x – 528 = 0, which is the required representation of the problem mathematically.

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

Let us consider,

The first integer number = x

Thus, the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306

⇒ x^{2 }+ x = 306

⇒ x^{2 }+ x – 306 = 0

∴ the two integers x and x+1, satisfies the quadratic equation, x^{2 }+ x – 306 = 0, which is the required representation of the problem mathematically.

**(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.**

Let us consider,

Age of Rohan’s = x years

∴ as per the given question,

Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

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

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 32x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

∴ the age of Rohan and his mother, satisfies the quadratic equation, x^{2} + 32x – 273 = 0, which is the required representation of the problem mathematically.

**(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**

Let us consider,

The speed of train = x km/h

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

As per second condition, the speed of train = (x – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

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

As we know,

Speed × Time = Distance

∴ (x – 8)(480/(x + 3) = 480

⇒ 480 + 3x – 3840/x – 24 = 480

⇒ 3x – 3840/x = 24

⇒ 3x^{2 }– 8x – 1280 = 0

∴ the speed of the train, satisfies the quadratic equation, 3x^{2 }– 8x – 1280 = 0, which is the required representation of the problem mathematically.