Solve the inequality a(x - 4)^2 + b < 10 for x, treating a and b as parameters.

Step-by-step solution

1. 1

   Isolate the squared term. Subtract $b$ from both sides:

   $a(x-4)^2<10-b$

   Nothing can be divided by $a$ yet, because $a$ might be zero or negative — that is precisely what forces a case analysis.

2. 2

   Case $a=0$ first, since it removes $x$ entirely. The inequality collapses to the parameter statement $b<10$, which does not involve $x$ at all:

   $b<10\Rightarrow x\in\mathbb{R};\qquad b\ge 10\Rightarrow x\in\varnothing$

3. 3

   Case $a>0$: divide without flipping.

   $(x-4)^2<\frac{10-b}{a}$

   If $b\ge 10$ the right side is $\le 0$, and a square can never be strictly less than a non-positive number, so the set is empty. If $b<10$ the right side is positive and $|x-4|<\sqrt{\tfrac{10-b}{a}}$, giving the bounded interval

   $4-\sqrt{\frac{10-b}{a}}<x<4+\sqrt{\frac{10-b}{a}}$

4. 4

   Case $a<0$: divide and flip.

   $(x-4)^2>\frac{10-b}{a}$

   Watch the sign of the quotient: with $a<0$, $b<10$ makes $\tfrac{10-b}{a}$ negative, so the inequality holds for every $x$ and the answer is $\mathbb{R}$. If $b=10$ the right side is $0$ and $(x-4)^2>0$ excludes only $x=4$.

5. 5

   Case $a<0$ with $b>10$: two rays. Now $\tfrac{10-b}{a}>0$ (negative over negative), so writing $c=\sqrt{\tfrac{10-b}{a}}$ gives $|x-4|>c$:

   $x<4-c\quad\text{or}\quad x>4+c$

6. 6

   Spot-check two of the seven branches numerically. Take $a=2$, $b=2$: the interval should be $4\pm\sqrt{4}=(2,6)$, and indeed at $x=3$, $2(1)+2=4<10$ $\checkmark$ while at $x=7$, $2(9)+2=20\not<10$ $\checkmark$. Take $a=-1$, $b=19$: $c=\sqrt{9}=3$, predicting $x<1$ or $x>7$; at $x=0$, $-16+19=3<10$ $\checkmark$ and at $x=4$, $19\not<10$ $\checkmark$.