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**Class X,
Solutions of NCERT (CBSE) Mathematics **

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*Chapter 2, Polynomials *

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__NCERT solutions for Maths Textbook Exercise 2.2__

(Page 33)

**Q 1: Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.**

**(i) x**

^{2}– 2x – 8 (ii) 4s^{2}– 4s + 1 (iii) 6x^{2}– 3 – 7x**(iv) 4u**

^{2}+ 8u (v) t^{2}– 15 (vi) 3x^{2}– x – 4__Solution__:

(i) x

^{2}– 2x – 8
For zeros of the polynomial,

x

^{2}– 2x – 8 = 0
or, x

^{2}– 4x + 2x – 8 = 0
or, x(x – 4) + 2(x – 4) = 0

or, (x – 4) (x + 2) = 0

Therefore, the zeros of (x

^{2}– 2x – 8) are 4 and –2.
Now, comparing the quadratic polynomial
with ax

^{2}+ bx + c, we get,
a = 1, b = –2 and c = –8. ………..(Eq 1)

So, we take α = 4 and β = –2.

Now, Sum of zeros = α + β = 4 – 2 = 2 =
– (–2)/1 = – b/a (after putting the values from Eq 1 above)

Product of zeros = αβ = 4.(– 2) = – 8 =
–8/1 = c/a.

Hence, verified.

(ii) 4s

^{2}– 4s + 1
For zeros of the polynomial,

4s

^{2}– 4s + 1 = 0
or, 4s

^{2}– 2s – 2s + 1 = 0
or, 2s(2s – 1) – 1(2s – 1) = 0

or, (2s – 1) (2s – 1) = 0

Therefore, the zeros of (4s

^{2}– 4s + 1) are 1/2 and 1/2.
Now, comparing the quadratic polynomial
with ax

^{2}+ bx + c, we get,
a = 4, b = –4 and c = 1. ………..(Eq 1)

So, we take α = 1/2 and β = 1/2.

Now, sum of zeros = α + β = ½ + ½ = 1 =
– (–4)/4 = – b/a (after putting the values from Eq 1 above)

Product of zeros = αβ =1/2 x ½ = ¼ =
c/a

Hence,
verified.

(iii) For zeros of the polynomial,

6x

^{2}– 3 – 7x = 0
or, 6x

^{2}– 9x + 2x – 3 = 0
or, 3x(2x – 3)
+ 1(2x – 3) = 0

or, (2x – 3)
(3x + 1) = 0

Therefore, the
zeros of 6x

^{2}– 3 – 7x are 3/2 and -1/3.
Now, comparing the quadratic polynomial
with ax

^{2}+ bx + c, we get,
a = 6, b = -7,
c = -3.

So, we take α = 3/2 and β = -1/3.

Now, sum of zeros = α + β = 7/6 = – (–7)/6
= – b/a (after putting the values from the expression above)

Product of zeros = αβ = –1/2 = –3/6 =
c/a

Hence,
verified.

(iv) Do it yourself
taking hints from the above solutions.

(v) For zeros of the polynomial,

t

^{2}– 15 = 0
or, (t + √15)
(t – √15) = 0or, t = ±√15

Therefore, the
zeros of (t

^{2}– 15) are –√15 and √15.
Now, comparing the quadratic polynomial
with ax

^{2}+ bx + c, we get,
t

^{2}+ 0.t – 15
So, a = 1, b =
0, c = –15

So, we take α = –√15 and β = √15.

Now, Sum of zeros = α + β = – √15 + √15
= 0 = 0/1 = – b/a

Product of zeros = αβ = – √15 . √15 = –
15 = –15/1 = c/a

Hence,
verified.

(vi) Do it yourself
taking hints from the above solutions.

**Q 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.**

__Solution__: (i) Let the quadratic polynomial be ax

^{2}+ bx + c, and its zeroes be α and β. We have,

Here we find, a
= 4, b = –1 and c = –4

So one
quadratic polynomial which fits the given condition is

4x

^{2 }– x – 4.__Solution__: (ii) Let the quadratic polynomial be ax

^{2}+ bx + c, and its zeroes be α and β. We have,

If, a = 3, b =
3√2, and c = 1, then one quadratic polynomial which fits the given condition is

3x

^{2}– 3√2x + 1.__Solution__: (iii) Let the quadratic polynomial be ax

^{2}+ bx + c, and its zeroes be α and β. We have,

α + β = 0 = 0/1
= –b/a, and

αβ = √5 = √5/1
= c/a

If a = 1, b =
0, c = √5 then, we can write a quadratic polynomial which fits the given
condition as

x

^{2}+ √5__Solution__: (iv) Consider a quadratic polynomial ax

^{2}+ bx + c, and whose zeroes be α and β. We have,

α + β = 1 = –(–1/1)
= –b/a, and

αβ = 1 = 1/1 =
c/a

If, a = 1, b = –1,
and c = 1, then one quadratic polynomial which fits the given condition can be

x

^{2}– x + 1__Solution__: (v) Try yourself taking cue from the above given solutions.

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