The quadratic equation $ax^2+8x+c=0$ has exactly one solution. If $a+c=10$, and $a<c$ find the ordered pair $(a,c)$.
To find the values of $a$ and $c$, I need to use the fact that the equation has exactly one solution. This means that the discriminant, which is the part under the square root in the quadratic formula, must be equal to zero. The discriminant is $8^2-4ac$, so I can write an equation: $8^2-4ac=0$. Simplifying, I get $4ac=64$, or $ac=16$. I also have the equation $a+c=10$, which I can use to eliminate one of the variables. I can try to solve for $a$ in terms of $c$: $a=10-c$. Then I can substitute this into the equation $ac=16$: $(10-c)c=16$. Expanding and rearranging, I get $c^2-10c+16=0$. This is another quadratic equation, which I can factor: $(c-8)(c-2)=0$. The solutions are $c=8$ and $c=2$. To find the corresponding values of $a$, I can plug these back into the equation $a=10-c$. I get $a=2$ when $c=8$, and $a=8$ when $c=2$. However, I have to remember the condition that $a<c$, which eliminates one of the possibilities. The only pair $(a,c)$ that satisfies this condition is $(a,c)=(2,8)$. 