Rooted

To get the answer, you can simply use the quadratic formula. In the quadratic formula, a = 1, b = -b, and c = b.

Since there is only 1 real root, the square root part of the quadratic formula must be equal to 0.

So, √b^2-4ac becomes √b^2-4b, which equals 0.

Factorising the terms:

√b(b-4) = 0

Now, square both sides:

b(b-4) = 0

Now, it is obvious that b can be either 0 or 4, so the answer is 0 + 4 = 4.

Of course, if you are ambitious you can plug in the values of b in the quadratic formula to find x.

Let $r$ be the one real root. Then $x^2 - bx + b = (x-r)^2 = x^2 -2rx + r^2$ . So we have $b=r^2$ and $b=2r$ . Combining these equations yields $\begin{aligned} r^2 &= 2r \\ r^2 - 2r &= 0 \\ r(r - 2) &= 0 \end{aligned}$ Thus $r=0$ , in which case $b=0$ , or $r=2$ , in which case $b=4$ . So the sum of all possible $b$ is $0 + 4 = \boxed{4}$ .