POLYNOMIALS, Class 10 CBSE Maths - NCERT solutions – Mathematics Textbook Exercise 2.2


Class X, Solutions of NCERT (CBSE) Mathematics  

Chapter 2, Polynomials


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) x2 – 2x – 8  (ii) 4s2 – 4s + 1  (iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u   (v) t2 – 15   (vi) 3x2 – x – 4
Solution:
(i) x2 – 2x – 8
For zeros of the polynomial,
x2 – 2x – 8 = 0
or, x2 – 4x + 2x – 8 = 0
or, x(x – 4) + 2(x – 4) = 0
or, (x – 4) (x + 2) = 0
Therefore, the zeros of (x2 – 2x – 8) are 4 and –2.
Now, comparing the quadratic polynomial with ax2 + 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) 4s2 – 4s + 1 
For zeros of the polynomial,
4s2 – 4s + 1 = 0
or, 4s2 – 2s – 2s + 1 = 0
or, 2s(2s – 1) – 1(2s – 1) = 0
or, (2s – 1) (2s – 1) = 0
Therefore, the zeros of (4s2 – 4s + 1) are 1/2 and 1/2.
Now, comparing the quadratic polynomial with ax2 + 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,
6x2 – 3 – 7x = 0
or, 6x2 – 9x + 2x – 3 = 0
or, 3x(2x – 3) + 1(2x – 3) = 0
or, (2x – 3) (3x + 1) = 0
Therefore, the zeros of 6x2 – 3 – 7x are 3/2 and -1/3.  
Now, comparing the quadratic polynomial with ax2 + 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,
t2 – 15 = 0
or, (t + √­­­­­­15) (t – √15) = 0or, t = ±√15    
Therefore, the zeros of (t2 – 15) are –√15 and √15.
Now, comparing the quadratic polynomial with ax2 + bx + c, we get,
t2 + 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 ax2 + 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
4x2 – x – 4.

Solution: (ii) Let the quadratic polynomial be ax2 + 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
3x2 – 3√2x + 1.

Solution: (iii) Let the quadratic polynomial be ax2 + 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
x2 + √5
Solution: (iv) Consider a quadratic polynomial ax2 + 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
x2 – x + 1

Solution: (v) Try yourself taking cue from the above given solutions.
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2 comments:

Gajsa rathore said...

nice job and thankssssss..

thanks a lot

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