if the sum of the squares of zeroes of the polynomial 6x2+x+k is 25/36,find the value of k?
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compare given expression with ax²+bx+c
a=6,b=1,c=k
let zeroes of polynomials are α and β
sum of the zeroes =α+β= -b/a= -1/6 ----(1)
product of the roots =αβ= c/a = k/6----(2)
given sum of the squares of zeroes =25/36
α²+β²= 25/36---(3)
do the square both sides of (1)
(α+β)²=(-1/6)²
α²+β²+2αβ=1/36
25/36+2αβ=1/36[from (3)]
2αβ=1/36-25/36
2αβ=(1-25)/36
2αβ=-24/36
αβ= -24/2*36
αβ=-1/3---(4)
but (2) =(4)
k/6 = -1/3
k= (-1/3) *6
k=-2
a=6,b=1,c=k
let zeroes of polynomials are α and β
sum of the zeroes =α+β= -b/a= -1/6 ----(1)
product of the roots =αβ= c/a = k/6----(2)
given sum of the squares of zeroes =25/36
α²+β²= 25/36---(3)
do the square both sides of (1)
(α+β)²=(-1/6)²
α²+β²+2αβ=1/36
25/36+2αβ=1/36[from (3)]
2αβ=1/36-25/36
2αβ=(1-25)/36
2αβ=-24/36
αβ= -24/2*36
αβ=-1/3---(4)
but (2) =(4)
k/6 = -1/3
k= (-1/3) *6
k=-2
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