# 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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