**Class 10 Quadratic Equations NCERT solutions Maths**

**Chapter 4: Quadratic Equations**

## **TEXTBOOK’S EXERCISE 4.1 **

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

(i) (x + 1)^{2} = 2 (x – 3)

(ii) x^{2 }– 2x = (–2) (3–x)

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

(iv) (x – 3) (2x + 1) = x (x + 5)

(v) (2x – 1) (x – 3) = (x + 5) (x–1)

(vi) x^{2} + 3x +
1 = (x – 2)^{2 }

(vii) (x + 2)^{3} = 2x(x^{2} – 1)

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

**Sol. **Standard form of quadratic equation is ax^{2} + bx + c = 0; a ≠ 0.

(i) (x + 1)^{2}
= 2(x – 3)

⇒ x^{2} + 1
+ 2x = 2(x – 3)

⇒ x^{2} + 1
+ 2x = 2x – 6

⇒ x^{2} + 1
+ 2x – 2x + 6 = 0 ⇒ x^{2} + 7 = 0

It is of the form ax^{2} + bx + c = 0,

Where a = 1, b = 0 and c = 7.

Hence, it is a quadratic equation.

(ii) x^{2} – 2x =
(–2) (3 – x)

⇒ x^{2} – 2x = – 6 + 2x

⇒ x^{2} – 2x – 2x + 6 = 0

⇒ x^{2} – 4x + 6 = 0

It is of the form ax^{2} + bx + c = 0,

where a = 1, b = – 4 and c = 6. Hence, it is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

⇒ x(x + 1) –2(x + 1) = x(x + 3) –1(x + 3)

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

⇒ x^{2} – x – 2 = x^{2} + 2x –
3

⇒ x^{2} + 2x – 3 – x^{2} + x +
2 = 0

⇒ 3x – 1 = 0

It is not of the
form ax^{2} + bx + c = 0.

Hence, it is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)

⇒ x(2x + 1) – 3(2x + 1) = x^{2} + 5x

⇒ 2x^{2} + x – 6x – 3 = x^{2}
+ 5x

⇒ 2x^{2} –
5x – 3 – x^{2} – 5x = 0

⇒ x^{2} –
10x – 3 = 0

It is of the form ax^{2} + bx + c = 0, where a = 1, b = –10 and c = –3. Hence, it is a quadratic equation.

#### Class 10 Quadratic Equations NCERT solutions Maths

It is of the form ax^{2} + bx + c = 0, where a =
1, b = –11 and c = 8.

Hence, it is a quadratic equation.

It is not of the form ax2 + bx + c = 0. Hence, the given equation is not a quadratic equation.

It is not of the form ax^{2} + bx + c = 0. Hence, the given equation is not a quadratic equation.

#### Class 10 Quadratic Equations NCERT solutions Maths

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.

(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.

**Sol.
**(i) Let breadth of
rectangular plot = x m

Then, length of the rectangular plot

= (2x + 1) m

So, area of rectangular plot

= [x (2x + 1)] m^{2} = (2x^{2 }+ x) m^{2}

As per condition :

Hence, given problem in the form of quadratic equation is

(ii) Let two consecutive positive integers are x and x + 1.

Product of integers = x (x + 1) = x^{2} + x As per condition:

Hence, given problem
in the form of quadratic equation is x^{2}
+ x – 306 = 0.

Hence, given problem
in the form of quadratic equation is x^{2}
+ x – 306 = 0.

(iii) Let present age of Rohan = x years

Present age of Rohan’s mother = (x + 26) years

After 3 years, Rohan’s age = (x + 3) years

After 3 years, Rohan’s mother’s age

= (x + 26 + 3) years = (x + 29) years

So, product = (x + 3) (x + 29)

= x^{2} + 29x + 3x + 87

= x^{2} + 32x + 87

As per condition: x^{2} + 32x + 87 = 360

⇒ x^{2} +
32x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

Hence, given problem in the form of quadratic equation is x^{2} + 32x – 273 = 0.

(iv) Let speed of train = x km/hour

Distance covered by train = 480 km

Time taken by train = Distance/ Speed = 480 /x hour

If speed of train is decreased by 8 km/hour

Then, new speed of train = (x – 8) km/hour Then, Time taken by train = 480

As per condition:

⇒ 3840 = 3(x^{2} – 8x) ⇒ 3x^{2
}– 24x = 3840

⇒ x^{2}
– 8x = 1280

⇒ x^{2}
– 8x – 1280 = 0

Hence, given problem in the form of quadratic equation is

x^{2} – 8x – 1280 = 0.