Question

If the sum of the squares of zeroes of the polynomial 6x^2+x+k is 25/36. Find the value of k.

Answer 1

$\bf\large\underline\pink{Solution:-}$

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

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Answer 2

Answer:

Sum of squares of zeroes=25/36

sum of zeroes=-b/a

Product of zeroes,alphabeta=c/a

-b/a=-1/6

c/a=k/6 (1)

square sum of zeroes =-1/6

sum of zeroes=-1/6

(alpha+beta)^2=(-1/6)^2

alpha^2+2alphabeta+beta^2=1/36

25/36+2alphabeta=1/36

2alphabeta=1/36-25/36

2alphabeta=-24/36

alphabeta=-24/36*2

alphabeta=-1/3

c/a=-1/3

k/6=-1/3

k=-2