The $x^2+ax+b = 0$ has only one solution and this is $x = -1+1/2$. What are the values of $a$ and $b$? This is a two part question. I was able to solve the first part. I need help with the second part.
a) The equation $x^2 + ax + b = 0$ has solutions $x = 2$ and $x = -5$. Find $a$ and $b$.
I was able to solve this one. we just have to set $f(2)$ equal to $f(-5)$. We will get $a=3$ and $b=-10$
b)  The $x^2+ax+b = 0$ has only one solution and this is $x = -1+1/2$.
What are the values of $a$ and $b$?
I don't understand how a quadratic equation can have only one solution. Isn't a quadratic equation always supposed to have two solutions. How do I solve this when only one solution is given? What do I set the expression equal to?
 A: The roots of a polynomial can be repeated. This happens when the roots at same which are of the form $(x-c)^2=x^2-2cx+c^2$.
A: If $x^2+ax+b=0$ has one solution, then the quadratic has a repeated root. Both of the roots could be equal to $5$, for instance. If the roots are repeated, then the discriminant $\Delta$ must equal $0$. Here, this means that
$$
\Delta = a^2-4b=0
$$
and so $4b=a^2$. Hence,
$$
x^2+ax+\frac{a^2}{4}=\left(x+\frac{a}{2}\right)^2=0 \, ,
$$
Can you work things out from here?
A: *

*

$$\begin{align}&x^2+ax+b \\
\iff &(2x+a)^2=a^2-4b\end{align}$$
If the quadratic equation has one real root, then $\Delta=a^2-4b=0$.
This implies,
$$\begin{cases}2x+a=0 \\a^2=4b \end{cases}\implies \begin{cases} a=-2x \\ b=x^2\end{cases}$$



*

If the quadratic has one real root, then, $x_1=x_2$.
By the Vieta's formulas, we have
$$\begin{cases} x_1+x_2=-a \\x_1x_2=b \end{cases} \implies  \begin{cases} a=-2x \\ b=x^2\end{cases}$$



*

If the quadratic has one real root, then
$$x^2+ax+b=(x-x_1)^2$$
$$\begin{align}&\implies x^2+ax+b=x^2-2xx_1+x_1^2 \\
&\implies a=-2x_1,~ b=x_1^2.\end{align}$$
