Question

Find the domain and solve the inequalities:
12/x^2+2x<3/x^2+4x+4

Answer 1

The domain and solution of the inequality is $(-\frac{8}{3}, -2)U(-2,0)$.

$\frac{12}{x^2+2x}<\frac{3}{x^2+4x+4}$

$\frac{4}{x(x+2)}<\frac{1}{(x+2)^2}$

$\frac{4}{x(x+2)}-\frac{1}{(x+2)^2}<0$

$\frac{1}{x+2}[\frac{4}{x}-\frac{1}{(x+2)}]<0$

$\frac{1}{x+2}[\frac{4x+8-x}{x(x+2)}]<0$

$\frac{1}{x+2}\times \frac{3x+8}{x(x+2)}<0$

Critical points are $x=-\frac{8}{3}, 0, -2$

Now, divide the number line in $4$ intervals and check for the intervals where the inequality is true.

So, Domain $= (-\frac{8}{3}, -2)U(-2,0)$.

Learn more about solving inequalities here:

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