Question

x²+3|x|+2>0 solve for X in integral​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm :\longmapsto\: {x}^{2} + 3 |x| + 2 > 0$

We know,

$\underbrace{ \boxed{ \bf \: { |x| }^{2} = {x}^{2}}}$

So, given inequality can be rewritten as

$\rm :\longmapsto\: { |x| }^{2} + 3 |x| + 2 > 0$

Let assume that | x | = y,

So, given inequality reduced to

$\rm :\longmapsto\: {y}^{2} + 3y + 2 > 0$

$\rm :\longmapsto\: {y}^{2} + 2y + y + 2 > 0$

$\rm :\longmapsto\:y(y + 2) + 1(y + 2) > 0$

$\rm :\longmapsto\:(y + 2)(y + 1) > 0$

We know that,

if a and b are two positive real numbers such that a < b and ( x - a ) ( x - b ) > 0 then x < a or x > b.

So, using this

$\rm :\implies\:y < - 2 \: \: \: or \: \: \: y > - 1$

Consider,

$\rm :\longmapsto\:y < - 2$

$\rm :\longmapsto\: |x| < - 2$

which is not possible.

Hence , there is no solution in this case.

Now, Consider,

$\rm :\longmapsto\:y > - 1$

$\rm :\longmapsto\: |x| > - 1$

$\bf\implies \: |x| \geqslant 0$

$\bf\implies \:x \: \in \: R$

But it is given that x is an integer.

So,

$\bf\implies \:x \: \in \: \{. \: . \: ., \: - 3, - 2, - 1,0,1,2,3, \: . \: . \: . \}$

Additional Information :-

$\rm :\longmapsto\: |x| < y\rm \implies\: - y < x < y$

$\rm :\longmapsto\: |x| \leqslant y\rm \implies\: - y \leqslant x \leqslant y$

$\rm :\longmapsto\: |x| > y\rm \implies\: x < - y \: \: or \: x > y$

$\rm :\longmapsto\: |x| \geqslant y\rm \implies\: x \leqslant - y \: \: or \: x \geqslant y$

$\rm :\longmapsto\: |x - z| < y\rm \implies\: z- y < x <z + y$

$\rm :\longmapsto\: |x - z| \leqslant y\rm \implies\: z- y \leqslant x \leqslant z + y$