real root (1 Viewer)

[juantheron]

find value of for which the equation has one real solution

 

[RealiseNothing]

Don't you just let the discriminant equal 0?

So

Solving for k gives

But I don't think you can do this since the quadratic is can you?

 

[funnytomato]

it's an even function ,i.e. f(alpha)=f(-alpha)
if alpha is a real zero, so is -alpha

there is only one real solution -----> then it must be x=0, substitue and we get k^2-1=0 , k= 1 or -1

these two values of k would give x^4 - x^2=0 and x^4 + 3x^2=0 respectively

the 1st eqn has 3 real solns
2nd one has 1 real soln, hence k=-1

 

[juantheron]

Thanks friends, yes answer is k=-1