What can be the value of $k$ for which the equation $9x^2+2kx-1=0$ has real roots?
Thing should be known
When the quadratic equation has real roots, then $d=b^2-4ac \ge 0$ .
Where $a$, $b$ and $c$ are the constant terms of a quadratic equation $ax^2+bx+c=0$.
On
If $b^2 - 4ac \geq 0$, then this gives you $4k^2 + 36 \geq 0$. What can you say about $k$ from here, considering that all the numbers appearing in the inequality are non-negative?
In general, what can you say about the roots of a quadratic $ax^2 + bx + c$, where $a > 0$ and $c \leq 0$ or where $a < 0$ and $c \geq 0$?
Besides to @Rankeya's answer, you can see that: $$9x^2+2kx=1\Leftrightarrow 9x^2+2kx+\frac{k^2}{9}=1+\frac{k^2}{9}\\\Leftrightarrow \left(3x+\frac{k}{3}\right)^2=\frac{k^2}{9}+1\Leftrightarrow 3x+\frac{k}{3}=\pm\sqrt{\frac{k^2}{9}+1}$$ or $$3x=\pm\sqrt{\frac{k^2}{9}+1}-\frac{k}{3}$$ which shows that you always have two distinct solutions for $x$ for all $k$.