Find value of $k$ for which the equation has real roots What can be the value of $k$ for which the equation $9x^2+2kx-1=0$ has real roots?

Thing should be known

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*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 $ax^2+bx+c=0$.


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 $k$.
