# Investigating the solutions of $9^{x}+k(3^{x})+2=0$

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

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.