Question

find the value of k for which one root of the equation kx2-14x+8=0 is six times the other

Answer 1

let one root be α and the other be 6α
In p(x)=kx²-14x+8=0
then
Sum of zeroes=-b/a
               α+6α=-(-14)k
                   7α=14/k
                     α=2/k
Product of zeroes=c/a
                            6α×α=8/k
                               6α²=8/k
                          6(2/k)²=8/k
                              24/8=k²/k ⇒ k=3

:) Hope this ans would help u...

Answer 2

Let the one root be α

And the other root be 6α

Now the given equation is kx^2-14x+8

Sum of two roots =-b/a
=>α+6α=14/k
=>7α=14/k
=>α=14/7k
=>α=2k

Now,
The product of two roots =c/a
=>α.6α=8/k
=>6α^2=8/k........¡

Putting α=2k in ¡ we get
=>6×2/k×2/k=8/k 24/k=8/k K=3