Question 933103
{{{kx=x^2-3x+5}}} by substitution.
{{{x^2-3x+5-kx=0}}}
{{{x^2-3x-kx+5=0}}}
{{{x^2-(3+k)x+5=0}}}


The expression for that equation's discriminant is {{{(3+k)^2-4*1*5}}}.
Discriminant value of 0 means the quadratic equation {{{x^2-(3+k)x+5=0}}} is a perfect square on the left side member, the quadratic expression.


{{{(3+k)^2-20=0}}}
{{{k^2+6k+9-20=0}}}
{{{k^2+6k-11=0}}}
{{{k=(-6+- sqrt(36+44))/2}}}
{{{k=(-6+- sqrt(80))/2}}}
{{{k=(-6+- 4*sqrt(5))/2}}}
{{{highlight(k=-3+- 2sqrt(5))}}}, either of these two values for k so that the two given equations meet in exactly one point.