The solution set of a(x - 4) squared plus b, less than 10, is k - 8 < x < 2k + 1. Find k.

Step-by-step solution

1. 1

   Locate the axis of symmetry. The expression $a(x-4)^2+b$ is in vertex form: its graph is a parabola whose vertex sits at $x=4$, and it is symmetric about the vertical line $x=4$. Nothing about $a$ or $b$ changes that line.

2. 2

   Argue that the interval must be centred on the axis. The solution set is bounded (an interval, not a union of two rays), which forces $a>0$ — an upward parabola dips below the horizontal line $y=10$ on a single interval. Its two boundary points are where $a(x-4)^2+b=10$, i.e. $(x-4)^2=\frac{10-b}{a}$, so they are $x=4\pm\sqrt{\tfrac{10-b}{a}}$: equidistant from 4. Hence the midpoint of the solution interval is exactly $4$.

3. 3

   Set the midpoint of the given interval equal to 4.

   $\frac{(k-8)+(2k+1)}{2}=4\quad\Rightarrow\quad \frac{3k-7}{2}=4\quad\Rightarrow\quad 3k-7=8$

4. 4

   Solve for $k$.

   $3k=15\quad\Rightarrow\quad k=5$

5. 5

   Verify the interval and check it is nondegenerate. With $k=5$ the endpoints are $k-8=-3$ and $2k+1=11$, so the solution set is $-3<x<11$; its midpoint is $\frac{-3+11}{2}=4$ ✓, matching the axis. The interval is also genuinely nonempty ($-3<11$), as a valid solution set must be.

6. 6

   Note what the data does and does not pin down. The half-width is $11-4=7$, so $49a+b=10$, i.e. $b=10-49a$. Any $a>0$ paired with that $b$ produces the same interval — so $k=5$ is uniquely determined even though $a$ and $b$ individually are not.