If the product of the roots of the equation #x^2-3kx+2e^(2log_e(k))-1=0#, then the value of #k# is?

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Answer 1 · 2018-07-31T06:29:09

#color(blue)(alpha*beta=c/a=(2k^2-1)/1=2k^2-1#

Here ,

#x^2-3kx+2e^(2log_e (k))-1=0#

Comparing with quadratic equation #ax^2+bx+c=0#

#color(red)(a=1 ,b=-3k and#

# c=2e^(2log_e (k))-1=2e^(log_e(k^2))-1to[becauselnX^n=nlnX]#

#:.color(red)(c=2(k^2)-1)......................to[because e^(lnX)=X]#

Let , #alpha and beta# be the roots of the equation.

So, the product of the roots of eqn. is :

#color(blue)(alpha*beta=c/a=(2k^2-1)/1=2k^2-1#