I have worked this one through but still not 100% sure.

  1. the discriminant is $D=(k-2\sqrt{2})(k+2\sqrt{2})$.
  2. the quadratic equation gives $3^{x}=\dfrac{-k\pm\sqrt{k^2-8}}{2}$.
  3. as the RHS must be at least $0$ for this equation to have any solutions I did some work using inequalities

I concluded that this equation:

  • cannot have repeated roots when $k=-2\sqrt{2}$

  • has two distinct roots when $k<-2\sqrt{2}$

But not too sure about the details for the case when the equation has two distinct roots. How would you determine this?

Thank you


1 Answer 1


To be able to compute the roots of $3^{2x} + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 \geq 0$. So $k\leq -2\sqrt{2}$ or $k\geq 2\sqrt{2}$. In case of either equality, you have zero discriminant.

You must also meet the condition $$-k \pm \sqrt{k^2-8} > 0. $$ This leaves you with $k < -2\sqrt{2}$.

Solving analytically the inequality $-k \pm \sqrt{k^2-8} > 0$ can be prone to mistakes so make sure to double check.

When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = \sqrt{x^2-8}$ and $h(x) = -g(x)$ respectively.

enter image description here

  • $\begingroup$ Thank you. This makes sense now. $\endgroup$
    – Will Kim
    Aug 28, 2018 at 15:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .