Question

Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.

Answer 1

Answer:

$x=-1, S=\{x\:\in\mathbb{R}:x=-1\}$

Step-by-step explanation:

For the sake of clarity, assuming you meant:

$|x^{2}+2x+3|\le2$

1) Absolute Value or Modulus functions has one property that assures us that:

$If |a|$

2) Solving for x

$I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}$

Not defined in Real Set of Numbers

Evaluating $x^{2}+2x+3\leqslant 2$:

$x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}$

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for $x^{2}+2x+3\geqslant -2\: \in \mathbb{C}$. The only solution in the Real Set for the inequality $|x^{2}+2x+3|\le2$ is $S=\{-1\}$, i.e. x=-1.