Question

Find the value of a for which the solution of the inequality is all real numbers.

Answer 1

Answer:

$a=3,\text{ then } x\in(-\infty,\infty)\\ \\a\in(-\infty,3),\text{ then }x\in \left[\dfrac{-4a-8}{3-a},\infty\right)\\ \\a\in(3,\infty),\text{ then }x\in \left(-\infty, \dfrac{-4a-8}{3-a}\right]$

Step-by-step explanation:

In the inequality

$3x+8+2ax\ge 3ax-4a\ \ \ (1)$

a is an arbitrary real number.

Separate the terms with x into left side and the terms without x in the right side:

$3x+2ax-3ax\ge -4a-8\\ \\3x-ax\ge -4a-8\\ \\(3-a)x\ge -4a-8\ \ \ (2)$

First, look at the leading coefficient at x. If this coefficient is equal to 0 (when $a=3$), then the inequality is

$0\ge -4a-8\\ \\0\ge -12-8\ [\text{Substituted }a=3]\\ \\0\ge -20$

This is true inequality for all x, so at $a=3,$ the inequality (1) has the solution $x\in (-\infty,\infty)$

Now, if the leading coefficient

$3-a>0\\ \\a$

then we can divide the inequality (2) by this positive number $3-a$ and get

$x\ge \dfrac{-4a-8}{3-a}$

If the leading coefficient

$3-a3,$

then we can divide the inequality (2) by this negative number $3-a$ and get

$x\le \dfrac{-4a-8}{3-a}$

So, the answer is

$a=3,\text{ then } x\in(-\infty,\infty)\\ \\a\in(-\infty,3),\text{ then }x\in \left[\dfrac{-4a-8}{3-a},\infty\right)\\ \\a\in(3,\infty),\text{ then }x\in \left(-\infty, \dfrac{-4a-8}{3-a}\right]$