**NCERT Solutions Maths Class 10 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) (2 x − 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}**

^{ }

**Solution: (i)** **( x + 1)^{2} = 2(x − 3)**

⇒ *x*^{2} + 2*x* + 1 =
2*x *– 6 [Since (*a*
+ *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}]

⇒ *x*^{2}
+ 7 = 0

Here,
the degree of the equation is 2.

Therefore,
it is a quadratic equation.

**(ii**) *x*^{2
}− 2*x* = (−2)(3 − *x*)

⇒ *x*^{2} – 2*x* = −6 + 2*x*

⇒ *x*^{2} – 2*x* − 2*x *+ 6 = 0

⇒ *x*^{2} – 4*x* + 6 = 0

Here,
the degree of the equation is 2.

Therefore,
it is a quadratic equation.

** **

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

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

⇒ *x*^{2 }+ *x* – 2*x*
– 2 – *x*^{2 }– 3*x* + *x*
+ 3 = 0

⇒ *x *− 2*x *– 2 − 3*x *+ *x *+ 3 = 0

⇒ −3*x *+ 1 = 0

Here,
the degree of the equation is 1.

Therefore,
it is not a quadratic equation.

**(iv) ( x − 3)(2x + 1)
= x(x + 5)**

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

⇒ 2*x*^{2 }+ *x* – 6*x*
– 3 – *x*^{2 }– 5*x* = 0

⇒ *x*^{2} – 10*x* – 3 = 0

Here,
the degree of the equation is 2.

Therefore,
it is a quadratic equation.

** **

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

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

⇒ 2x^{2 }– 7x
+ 3 – x^{2 }+ x – 5x + 5 = 0

⇒ *x*^{2 }− 11*x *+ 8 = 0

Here,
the degree of the equation is 2.

Therefore,
it is a quadratic equation.

** **

**(vi) x^{2}
+ 3x + 1 = (x − 2)^{2}**

⇒ *x*^{2 }+ 3*x*
+ 1 = *x*^{2} – 4*x*^{ }+
4 [ Since (*a*
– *b*)^{2}
= *a*^{2} –** **2*ab* + *b*^{2}]

⇒ *x*^{2 }+ 3*x* + 1 – *x*^{2
}+ 4*x* – 4 = 0

⇒ 7*x *– 3 = 0

Here,
the degree of the equation is 1.

Therefore,
it is not a quadratic equation.

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

⇒ *x*^{3 }+ 2^{3 }+ 3(*x*)(2)(*x* + 2) = 2*x*(*x*^{2}
– 1) [Since (*a*
+ *b*)^{3} = *a*^{3} + *b*^{3} + 3*ab*(*a* + *b*)]

⇒ *x*^{3 }+ 8 + 6*x*(*x* + 2) = 2*x*^{3} – 2*x*

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

⇒ 2*x*^{3 }– 2*x*
– *x*^{3 }– 8 – 6*x*^{2 }– 12*x* = 0

⇒ *x*^{3 }– 6*x*^{2
}– 14*x* – 8 = 0

Here,
the degree of the equation is 3.

Therefore,
it is not a quadratic equation.

` `

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

⇒ *x*^{3 }– 4*x*^{2
}– *x* + 1 = *x*^{3 }– 2^{3 }– 3(*x*)(2)(*x *− 2) [Since (*a*
– *b*)^{3} = *a*^{3} – *b*^{3} – 3*ab*(*a* – *b*)]

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

⇒ 2*x*^{2 }– 13*x* + 9 = 0

Here,
the degree of the equation is 2.

Therefore,
it is a 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.**

**(ii) The product of two consecutive numbers 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)** It is given that the area of a
rectangular plot is 528 m^{2}.

Let
the breadth of the rectangular plot be *x*
metres.

Then, according to the given condition, length
= (2*x *+ 1) metres

Area
of a rectangle = length × breadth

⇒ 528 = (2*x *+ 1)*x*

⇒ 528 = 2*x*^{2 }+ *x*

⇒2*x*^{2} + *x* – 528 = 0

This
is the required quadratic equation.

**(ii)** Let the two consecutive positive
integers be *x* and (*x* + 1).

According to the given condition, we have

*x*(*x *+ 1) = 306

⇒ *x*^{2 }+ *x* = 306

⇒ *x*^{2 }+ *x* – 306 = 0

This
is the required quadratic equation.

** **

**(iii)** Let the present age of Rohan be *x* years.

Then,
the present age of Rohan’s mother = (*x*
+ 26) years

Age
of Rohan after 3 years = (*x* + 3)
years

Age
of Rohan’s mother after 3 years = *x* +
26 + 3 = (*x* + 29) years

According
to the given condition, we have

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

⇒ *x*^{2 }+ 29*x* + 3*x*
+ 87 = 360

⇒ *x*^{2 }+ 32*x*
– 273 = 0

This
is the required quadratic equation.

** **

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

Time taken by the train to cover 480 km = 480/*x* hours

If the speed had been 8 km/h less, then the
time taken would be 480/(*x *− 8) hours

According to the
given condition, if the speed had been 8 km/h less, then the time taken to
cover the same distance is 3 hours more.

Therefore, 480/(*x *− 8) = 480/*x *+ 3

⇒ 480/(*x *− 8) − 480/*x *= 3

⇒ 480[x − (*x *− 8)] = 3*x*(*x
*– 8)

⇒ 480(*x *– *x *+ 8) = 3*x*(*x *– 8)

⇒ 480 × 8 = 3*x*(*x *– 8)

⇒ 3840 = 3*x*^{2 }– 24*x*

⇒ 3*x*^{2 }– 24*x *– 3840 = 0

Dividing
the above equation by 3, we get

⇒ *x*^{2
}– 8*x *– 1280 = 0

This
is the required quadratic equation.