Find a non-zero value for $a$ such that $ax^2+8x+4=0$ has only one solution.
I know that a quadratic equation of the form $ax^2+bx+c=0$ has only one solution when the discriminant, $b^2-4ac$, is equal to zero. So in this case, I want to find $a$ such that $8^2-4a(4)=0$. Simplifying, I get $64-16a=0$, or $16a=64$. Dividing by 16, I get $a=4$. To check, I can plug in $a=4$ into the original equation and see if it has only one solution. I get $4x^2+8x+4=0$, which factors as $(2x+2)^2=0$. This means $x=-1$ is the only solution, as desired. 