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Homesquaring condition for square root inequality
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squaring condition for square root inequality
Of the inequalities $\sqrt{a}\lessgtr b$,

both are undefined when $a<0$;

both can be sidewise squared when $a\geqq 0$ and $b\geqq 0$;

$\sqrt{a}>b$ is identically true if $a\geqq 0$ and $b<0$.

$\sqrt{a}<b$ is identically untrue if $b<0$;
The above theorem may be utilised for solving inequalities involving square roots.
Example. Solve the inequality
$\displaystyle\sqrt{2x+3}\;>\;x.$  (1) 
The reality condition $2x+3\geqq 0$ requires that $x\geqq1\frac{1}{2}$. For using the theorem, we distinguish two cases according to the sign of the right hand side:
$1^{\circ}$: $1\frac{1}{2}\leqq x<0$. The inequality is identically true; we have for (1) the partial solution $1\frac{1}{2}\leqq x<0$.
$2^{\circ}$: $x\geqq 0$. Now we can square both sides, obtaining
$2x+3\;>\;x^{2}$ 
$x^{2}2x3\;<\;0$ 
The zeros of $x^{2}\!\!2x\!\!3$ are $x=1\pm 2$, i.e. $1$ and $3$. Since the graph of the polynomial function is a parabola opening upwards, the polynomial attains its negative values when $1<x<3$ (see quadratic inequality). Thus we obtain for (1) the partial solution $0\leqq x<3$.
Combining both partial solutions we obtain the total solution
$1\frac{1}{2}\;\leqq\;x\;<\;3.$ 
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