Question:
find the zeroes of 4 root 3 x2 5 x - 2 root 3 and verify the relation between the zeroes and cofficient of polynomial.
Answer:
Given polynomial is p(x) = 4√3x2 + 5 x - 2√3 = 0 ............1
= 4√3x2 + 5 x - 2√3 = 0
= 4√3x2 + 8x - 3x - 2√3 = 0
= 4x(√3x + 2) - √3(√3x + 2) = 0
=>(√3x + 2)*(4x - √3 ) = 0
=> x = √3/4, -2/√3
So zeroes are √3/4, -2/√3
Sum of zeroes =(√3/4) + (-2/√3) = √3/4 + -2/√3 = (3-8)/4√3 = -5/4√3
Product of zeroes = (√3/4) * (-2/√3) = -(√3*2)/(4*√3) = -1/2
Again from equation 1
Sum of zeroes = -5/4√3
product of zeores = -2√3/4√3 = -1/2
Hense it is varified.
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