if one of the zeroes of polynomial x²-kx +16 is cube of other find the value of k
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Hello.
let's begin solving.
Let the two roots be a & a3
So, x²-kx +16=(x-a) (x-a3)
x²-kx +16= x²+(-a-a3)x+a4
on comparing the corresponding terms, we get
-kx=-(a+a3)x
k=a+a3
a4=16
a=2
k=2+23
k=10 (ans)
Hope it helps.
let's begin solving.
Let the two roots be a & a3
So, x²-kx +16=(x-a) (x-a3)
x²-kx +16= x²+(-a-a3)x+a4
on comparing the corresponding terms, we get
-kx=-(a+a3)x
k=a+a3
a4=16
a=2
k=2+23
k=10 (ans)
Hope it helps.
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