\section*{Problem 4 - 061034}

Determine all real numbers \(k\) for which the equation

\[
x^{2}+x+3=k\left(x^{2}+5\right)
\]

is a quadratic equation in \(x\) that

a) has a double root!

b) has two distinct real roots!
}

The given equation is quadratic in \(x\) if and only if \(k \neq 1\). It is then equivalent to the equation

\[
x^{2}+\frac{1}{1-k} x+\frac{3-5 k}{1-k}=0
\]

This equation has real solutions if and only if the following radicands are non-negative, and the solutions are

\[
x_{1,2}=\frac{1}{2(k-1)} \pm \sqrt{\frac{1}{4(1-k)^{2}}-\frac{3-5 k}{1-k}}=\frac{1 \pm \sqrt{-20 k^{2}+32 k-11}}{2(k-1)}
\]

a) The equation (1) has a double solution if and only if

\[
-20 k^{2}+32 k-11=0 \quad \text { i.e. } \quad k^{2}-\frac{8}{5} k+\frac{11}{20}=0
\]

This is the case for \(k=\frac{11}{10}\) and for \(k=\frac{1}{2}\) and only for these values.

b) The equation (1) has two distinct real solutions if and only if \(-20 k^{2}+32 k-11>0\). This implies

\[
k^{2}-\frac{8}{5} k+\frac{11}{20}<0
\]

From this, we get

\[
\frac{1}{2}<k<\frac{11}{10}
\]

There is no value of \(k\) that satisfies both \(k-\frac{11}{20}<0\) and \(k-\frac{11}{20}>0\) simultaneously.

Therefore, the equation (1) has distinct real solutions for \(\frac{1}{2}<k<1\) and \(1<k<\frac{11}{10}\) and only for these values of \(k\).