Question

lf alpha and beta are the zeros of the quadratic polynomial f(x)=kx ^2+4x+4 such that alpha^2+beta^2=24find the value of k

Answer 1

hey mate here is ur answer
Givenalpha and beta are the zeroes of the polynomial kx^2+4x+4
I 'll denote alpha by x and beta by y.
x + y = -b/a
x+y=-4/k.......eq 1
xy= c/a
xy= 4/k........eq 2
Squaring both sides of eq 1
(x+y)2= (-4/k)2
x2+y2+2xy=16/k2
24+2xy=16/k2 [from x2+y2 = 24]
Substituting the value of xy from eq2,
24+2*4/k=16/k2
24+8/k=16/k2
24k+8/k=16/k2
24k+8=(16/k2)*(k)
24k+8=16/k
24k2+8k=16
or 24k2+8k -16=0
3k2+k-2=0
3k2+3k-2k-2=0
3k(k+1)-2(k+1)=0
(3k-2)(k+1)=0
Therefore,k is either -1 or 2/3

Answer 2

alpha^2 + beta^2 = (alpha+beta)^2 - 2(alpha)(beta) 
Alpha + beta = -4/k
(alpha)(beta)= 4/k
alpha^2 + beta^2 = [(-4/k)^2 -2(4/k)]

                          24 = [16/k^2 - 8/k]                           24=[16 - 8k/k^2]                          
                           24k^2=16 -8k

24k^2 + 8k - 16 =0(k+1)(24k-16)=0
k=-1
k=16/24=2/3








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