10th CBSE math solution for exercise 2.2 part 1

This page 10th CBSE math solution for exercise 2.2 part 1is 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 part 1

(iii) 6 x² – 3 – 7 x 

Solution:

Find let us find the zeroes of the quadratic polynomial.

6 x² – 3 – 7 x = 0

6 x²  - 7 x – 3 = 0

 (3 x + 1) ( 2 x - 3) = 0

3 x + 1 = 0       2 x – 3 = 0

   3 x = -1            2 x = 3

      x = -1/3             x = 3/2

So the values of α = -1/3 and β = 3/2

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

6 x²  - 7 x – 3 = 0

ax² + b x + c = 0

a = 6   b = -7   c = -3

Sum of zeroes α + β = -b/a

              (-1/3) + (3/2) = -(-7)/6

                   (- 2 + 9)/6 = 7/6

                          7/6 = 7/6

Product of zeroes α β = c/a

                            (-1/3)(3/2) = -3/6

                              -1/2 = -1/2


(iv) 4 u² + 8 u 

Solution:

Find let us find the zeroes of the quadratic polynomial.

4 u² + 8 u = 0

4 u (u + 2) = 0

 4 u = 0       u + 2 = 0

   u = 0            u = -2

So the values of α = 0 and β = -2

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

4 u² + 8 u = 0

ax² + b x + c = 0

a = 4   b = 8   c = 0

Sum of zeroes α + β = -b/a

                            0 + (-2) = -8/4

                                  - 2 = -2

        Product of zeroes α β = c/a

                            (0)(-2) = 0/4

                               0 = 0


(v) t² - 15 

Solution:

Find let us find the zeroes of the quadratic polynomial.

t^2 - 15 = 0

t²  = 15

  t = √15

  t = ± √15

t = √15   t = - √15

So the values of α = √15 and β = -√15

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

t² - 15 = 0

ax² + b x + c = 0

a = 1   b = 0   c = -15

Sum of zeroes α + β = -b/a

                            √15 + (-√15) = -0/1

                                  0 = 0

        Product of zeroes α β = c/a

                            (√15)( -√15) = -15/1

                               -15 = -15

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