Question

find the value of k such that one zero of the polynomial 3x√2+(1+4k)x+k√2+5 is one third of the other​

Answer 1

Answer:

.k = 79/8

Step-by-step explanation:

Let a and a/3 are the roots of 3x^2+(1+4k).x+(k^2+5)=0.

a+a/3=-(1+4k)/3.

4a/3= -(1+4k)/3.

4a = -(1+4k).

a =-(1+4k)/4…………….(1)

a×a/3 =(k^2+5)/3.

a^2 = (k^2+5).

put a =-(1+4k)/4 from eq .(1).

{+(1+4k)^2}/16=(k^2+5).

1+16k^2+8k = 16k^2+80.

8k=79 => k = 79/8

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