# 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}  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  $x\geqq-1\frac{1}{2}$.  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 , obtaining

 $2x+3\;>\;x^{2}$
 $x^{2}-2x-3\;<\;0$

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 (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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