# 10th CBSE math solution for exercise 2.2 part 2

This page 10th CBSE math solution for exercise 2.2 part 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 part 2

(vi) 3 x² – x - 4

Solution:

Find let us find the zeroes of the quadratic polynomial.

3 x² – x - 4 = 0

(3 x - 4) (x + 1) = 0

(3 x - 4) = 0           (x + 1) = 0

3 x = 4                   x = -1

x = 4/3

x = 4/3   x = -1

So the values of α = 4/3 and β = -1

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

3 x² – x - 4 = 0

ax² + b x + c = 0

a = 3   b = -1   c = -4

Sum of zeroes α + β = -b/a

(4/3) + (-1) = -(-1)/3

(4-3)/3 = 1/3

1/3 = 1/3

Product of zeroes α β = c/a

(4/3)( -1) = -4/3

-4/3 = -4/3

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

(i) 1/4 ,-1

Solution:

α = 1/4

β = -1

General form of a quadratic equation

x² – (α + β) x + α β = 0

α + β = (¼) + 1

= (1 + 4)/4

= 5/4

α β = (1/4) ( 1)

= 1/4

x² – (5/4) x + (1/4) = 0

4 x² – 5 x + 1 = 0

(ii) √2,1/3

Solution:

α = √2

β = 1/3

General form of a quadratic equation

x² – (α + β) x + α β = 0

α + β = √2 + (1/3)

= (3√2 + 1)/3

α β = (√2) (1/3)

= √2/3

x² – [(3√2 + 1)/3] x + (√2/3) = 0

3 x² – (3√2 + 1) x + √2 = 0

(iii) 0, √5

Solution:

α = 0

β = √5

General form of a quadratic equation

x² – (α + β) x + α β = 0

α + β = 0 + √5

= √5

α β = (0) (√5)

= 0

x²– √5 x + (0) = 0

x² –√5  x = 0

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