Question 762·Medium·Math

Type $y=x^2-5x$ and $y=6$ into Desmos to graph each side of the given equation. At a crossing, the two sides have the same value, so its $x$-coordinate solves $x^2-5x=6$. Desmos shows two crossings.

Click both intersections of the parabola, the U-shaped graph, and the horizontal line. Desmos shows $(-1,6)$ and $(6,6)$. Their $x$-coordinates are different, so the distinct real solutions are $-1$ and $6$.

Type $-1+6$ into Desmos. It shows $5$, so add each distinct solution once. The sum of all distinct real solutions is $5$. Grid in 5.

The first route confirmed there are two distinct roots. For a quadratic in the form $ax^2+bx+c=0$, the root-sum relationship says: $r+s=-\frac{b}{a}$. Here $r$ and $s$ are the roots. The minus sign comes from expanding their factors: the number multiplying $x$ is $-a(r+s)$.

Subtract $6$ from both sides to put the equation in the form $ax^2+bx+c=0$: $x^2-5x-6=0$. In $x^2-5x-6=0$, the number multiplying $x^2$ is $a=1$, and the signed number multiplying $x$ is $b=-5$. Substitute them:

Reverse the sign of $-5$: $r+s=\frac{5}{1}$. Divide by $1$: $r+s=5$. So the distinct real solutions add to $5$. Grid in 5.