Question:
Find the zeroes of 3√2x^2 +13x +6√2 and verify the relation between the zeroes and coefficients of the polynomial
Answer:
Given, polynomial is: 3√2x2 + 13x + 6√2
= 3√2x2 + 13x + 6√2
= 3√2x2 + 9x + 4x + 6√2
= 3√2x2 + 9x + 2*√2*√2x + 6√2
= 3x(√2x + 3) + 2√2(√2x + 3)
= (√2x + 3)*(3x + 2√2)
So zeroes are -3/√2, -2√2/3
Sum of zeroes = (-3/√2) + (-2√2/3) = {(-9 -4)/3√2} = -13/3√2
Product of zeroes = (-3/√2) * (-2√2/3) = {-3 *(-2√2)}/{3*√2} = 6√2/3√2 = 6/3 = 2
Let a and b are zeroes of the given polynomial.
Sum of zeroes a + b = -13/3√2
product of zeores a*b = 6√2/3√2 = 6/3 = 2
Hense it is varified.
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