10th CBSE math solution for exercise 2.2

This page 10th CBSE math solution for exercise 2.2 is going to provide you solution for every problems that you find in the exercise no 2.2

10th CBSE math solution for exercise 2.2

(1) Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients

(i) x² – 2 x – 8

(ii) 4 s² – 4 s + 1 

(iii) 6 x² – 3 – 7 x     Solution

(iv) 4 u² + 8 u      Solution

(v) t² - 15        Solution

(vi) 3 x² – x - 4        Solution

(2) Find the quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 ,-1               Solution

(ii) √2,1/3        Solution

(iii) 0, √5        Solution

(iv)  1, 1        Solution

(v)  -1/4, 1/4        Solution

(vi) 4, 1        Solution

(i) x² – 2 x – 8

Solution:

Find let us find the zeroes of the quadratic polynomial.

x² – 2 x – 8 = 0

(x – 4) ( x + 2) = 0

x – 4 = 0          x + 2 = 0

  x = 4                   x = - 2

So the values of α = 4 and β = -2

Now we are going to verify the relationship between these zeroes and coefficients

x² – 2 x – 8 = 0

ax² + b x + c = 0

a = 1   b = -2   c = -8

Sum of zeroes α + β = -b/a

                     4 + (-2) = -(-2)/1

                          2 = 2

Product of zeroes α β = c/a

                            4(-2) = -8/1

                              -8 = -8


(ii) 4 s² – 4 s + 1 

Solution:

Find let us find the zeroes of the quadratic polynomial.

4 s² – 4 s + 1

(2 s – 1) ( 2 s - 1) = 0

2 s – 1 = 0    

  2 s = 1

   s =1/2   s = 1/2

So the values of α = 1/2 and β = 1/2

Now we are going to verify the relationship between these zeroes and coefficients

4 s² – 4 s + 1 = 0

ax² + b x + c = 0

a = 4   b = -4   c = 1

Sum of zeroes α + β = -b/a

              (1/2) + (1/2) = -(-4)/4

                          2/2 = 1

                          1 = 1

Product of zeroes α β = c/a

                            (1/2)(1/2) = 1/4

                              1/4 = 1/4