Problem 728

The equation $x^2+2kx+7k-10=0$ has two equal real roots precisely when the discriminant of this quadratic equation equals 0.

The discriminant, $\Delta$, equals $$\Delta = (2k)^2-4(1)(7k-10) = 4k^2 - 28k + 40$$ For the discriminant to equal 0, we have $4k^2-28k+40=0$ or $k^2-7k+10=0$ or $(k-2)(k-5)=0$.

Thus, $k=2$ or $k=5$.

We check that each of these values gives an equation with the desired property.

When $k=2$, the equation is $x^2+4x+4=0$ which is equivalent to $(x+2)^2=0$ and so only has one solution for $x$.

When $k=5$, the equation is $x^2+10x+25=0$ which is equivalent to $(x+5)^2=0$ and so only has one solution for $x$.

The sum of these values of $k$ is $2+5=7$.