1f one zero of polynomial ((k+1)x2-5x+ 5 is multiplicative inverse ofthe other, then l find the zeros of kx-3kx + 9 where k is constant.
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let zeroes of the polynomial be alpha
therefore,
multiplicative inverse be – 1/alpha
therefore, product of zeros = c/a
alpha×1/alpha= 5/k+1
k+1=5
k=5-1
k=4
putting this value of k in polynomial
kx²-3kx+9
(4)x²-3(4)x+9
4x²-12x+9
now,
4x²-12x+9=0
4x²-6x-6x+9=0
2x(2x-3)-3(2x-3)=0
(2x-3)(2x-3)=0
x=3/2, x=3/2.
therefore ,
alpha =3/2 and beta=3/2
therefore,
multiplicative inverse be – 1/alpha
therefore, product of zeros = c/a
alpha×1/alpha= 5/k+1
k+1=5
k=5-1
k=4
putting this value of k in polynomial
kx²-3kx+9
(4)x²-3(4)x+9
4x²-12x+9
now,
4x²-12x+9=0
4x²-6x-6x+9=0
2x(2x-3)-3(2x-3)=0
(2x-3)(2x-3)=0
x=3/2, x=3/2.
therefore ,
alpha =3/2 and beta=3/2
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