# SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 3.3 PART 3

This page Samacheer Kalvi Math Solution for Exercise 3.3 part 3 is going to provide you solution for every problems that you find in the exercise no 3.3

## Samacheer Kalvi Math Solution for Exercise 3.3 part 3

(vii) 2 x² - 2 2x + 1

Solution:

Let p(x) = 2 x² - 2 2x + 1  = (√2x - 1) (√2x - 1)

p(x) = 0 => (√2x - 1) (√2x - 1) = 0

√2x - 1 = 0

√2x = 1

x = 1/√2

x = 1/√2    x = 1/√2

p(1/√2) = 2 (1/√2)² - 2 2(1/√2) + 1

= 1 - 2 + 1

= 2 - 2

= 0

sum of zeroes = (1/√2) + (1/√2) = 2/√2

product of zeroes = (1/√2) (1/√2) = 1/2

2 x² - 2 2x + 1

ax² + bx + c

a = 2, b = -2 2 and c = 1

Sum of roots α + β = -b/a

α + β = -(-2 2)/2

α + β = 1

Product of roots α β = c/a

α β = 1/2

Thus the relationship verified

(viii) x² + 2 x - 143

Solution:

Let p(x) =  x² + 2 x - 143 = (x - 11) (x + 13)

p(x) = 0 => (x - 11) (x + 13) = 0

x - 11 = 0             x + 13 = 0

x = 11              x = -13

x = 11   x = -13

p(11) = (11)² + 2 (11) - 143

= 121 + 22 - 143

= 143 - 143

= 0

p(-13) = (-13)² + 2 (-13) - 143

= 169 - 26 - 143

= 143 - 143

= 0

sum of zeroes = 11 + (-13) = -2

product of zeroes = 11(-13) = -143

x² + 2 x - 143

ax² + bx + c

a = 1, b = 2 and c = -143

Sum of roots α + β = -b/a

α + β = 2/1

α + β = 2

Product of roots α β = c/a

α β = -143/1

= -143

Thus the relationship verified

In the page samacheer kalvi math solution for exercise 3.3 part 3 we are going to see the solution of next problem

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

(i) 3 , 1

Solution:

General form of quadratic equation with roots α and β is

x² - (α + β) x + αβ = 0

α = 3    β = 1

x² - (3 + 1) x + 3 (1) = 0

x² - 4 x + 3 = 0

(ii) 2 , 4

Solution:

General form of quadratic equation with roots α and β is

x² - (α + β) x + αβ = 0

α = 2    β = 4

x² - (2 + 4) x + 2 (4) = 0

x² - 6 x + 8 = 0 