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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