##
**Solutions of CBSE
Board, Class 10 Mathematics (SA-II / Term II) **

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*Chapter 4, Quadratic EquationS *

*Chapter 4, Quadratic EquationS*

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__NCERT Solutions
for Mathematics Textbook Exercise 4.1__

__NCERT Solutions for Mathematics Textbook Exercise 4.1__

*(Page 73)*

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}__Solution.1__:

(i) LHS = (x + 1)

^{2}= x^{2}+ 2x + 1
or, x

^{2}+ 2x + 1 = 2(x - 3)
or, x~~2x~~ + 1 = ~~2x~~ - 6

^{2}+
or, x

^{2}+ 7 = 0
The above expression is of the form: ax

^{2}+ bx + c = 0
Therefore, the given equation is a quadratic equation.

(ii) LHS = x

^{2}- 2x
Given that, x

^{2}- 2x = (–2) (3 - x)
or, x

^{2}- 2x - 2x + 6 = 0
or, x

^{2}- 4x + 6 = 0
It is of the form: ax

^{2}+ bx + c = 0
Therefore, the given equation is a quadratic equation.

(iii) LHS = (x - 2) (x + 1) = x(x + 1) - 2(x + 1)

or, LHS = x

^{2}- x -2,
Now as per given equation -

x

^{2}- x -2 = (x - 1) (x + 3) = x(x + 3) - 1(x + 3)
or, ~~x~~ - x -2 = ~~x~~
+ 3x - x - 3

^{2}

^{2}

or, –x - 2 - 2x + 3 = 0

or, –3x +1 = 0

It is not of the form: ax

^{2}+ bx + c = 0
Therefore, the given equation is not a quadratic
equation.

(iv), (v) & (vi) taking hints from above solutions
try top solve your-self.

(vii) LHS = (x +
2)

^{3}= x^{3}+ 2^{3}+ 3 (x) (2) (x + 2)
or, LHS = x

^{3}+ 8 + 6x^{2}+ 12x
As per the given equation,

x

^{3}+ 6x^{2}+ 12x + 8 = 2x(x^{2}- 1)
or, x

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

^{3}+ 6x^{2}+ 14x + 8 = 0
Since, it is not in the form of ax

^{2}+ bx + c = 0
Therefore, the given equation is not a quadratic
equation.

(viii) Do it yourself.

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 meters) 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/hr less, then it would have
taken 3 hrs more to cover the same distance. We need to find the speed of the
train.

__Solution.2__:

(i) Let the breadth be ‘x’.

As per the given condition,

Length = (2x + 1)

Now area of the rectangular plot = x (2x + 1) which is
given 528 m

^{2}
or, x (2x + 1) = 528

or, 2x

^{2}+ x - 528 = 0
Hence, the value of ‘x’ which satisfies the above
quadratic equation 2x

^{2}+ x - 528 = 0, is the breadth of the rectangular plot.
(ii) Let the first integer be ‘x’and the second integer
be (x + 1).

As per the given condition,

x (x + 1) = 306

or, x

^{2}+ x - 306 = 0
Hence, the value of ‘x’ which satisfies the above
quadratic equation: x

^{2}+ x - 306 = 0, where ‘x’ is the smallest integer.
(iii) Let Rohan’s age be ‘x’.

His mother’s age = (x + 26)

After 3 years,

Rohan’s age = x + 3

Mother’s age = x + 26 + 3 = x + 29

As per the given condition, the product of
their ages after 3 years is 360.

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

or, x

^{2}+ 32x + 87 = 360
or, x

^{2}+ 32x - 273 = 0
Hence, the value of ‘x’ which satisfies the above
quadratic equation: x

^{2}+ 32x - 273 = 0, is the present age of Rohan.
(iv) Taking hint
from above solutions try to solve this yourself.

At the end of completing all Exercises from
each Chapter given in

*NCERT Math textbook*, we shall provide**CBSE Notes (Hints) / CBSE Solutions, Mathematics Sample Questions with their Solutions**and many more…
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