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Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $

$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get

$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.

$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$

But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)

So we get no solution.

My Question is : Can we solve above question by using comparision of expressions?

Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$

If that way possible, please help me solve it. Thanks.

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  • $\begingroup$ Of course, proving inequality for all real $x$ amounts to disproving equality for all real $x$. $\endgroup$ Apr 5 '15 at 8:24
  • $\begingroup$ Vincenzo Oliva, would you like to explain me, Thanks $\endgroup$
    – juantheron
    Apr 5 '15 at 8:30
  • $\begingroup$ When squaring or cubing, extra roots are added and may be, as in your case, none of them is a solution. $\endgroup$ Apr 5 '15 at 8:30
  • $\begingroup$ @ClaudeLeibovici That was not what the OP is asking for. He did use the method of squaring and crosschecking all soultions found; he is asking for an alternative approach. $\endgroup$ Apr 5 '15 at 8:35
  • $\begingroup$ @HagenvonEitzen. I understood that and he did a good work as well as he asked a good question. $\endgroup$ Apr 5 '15 at 8:38
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For $x\ge1$ we have $$\sqrt{4x-1}\ge \sqrt {3x} $$ and $$\sqrt{x+1}\le \sqrt {2x}$$ hence $$\sqrt{x+1}-\sqrt{x-1}\le \sqrt{2x}<\sqrt{3x}\le\sqrt{4x-1} $$

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  • $\begingroup$ Well done! (+1). $\endgroup$ Apr 5 '15 at 8:40
  • $\begingroup$ SImple and nice way to do it ! When your answer appeared, I was about to propose something stupid since too complex. Thanks. $\endgroup$ Apr 5 '15 at 8:41
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Also, for $x\ge1$ $$\sqrt{4x-1}+\sqrt{x-1}\ge\sqrt{4x-1+x-1}=\sqrt{5x-2}>\sqrt{2x}\ge\sqrt{x+1}. $$

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Alternatively, isolating $ \sqrt{4x-1}$ and then multiplying both sides by $\sqrt{x+1}+\sqrt{x-1}$ makes it easier to conclude the LHS is smaller than the RHS: $$\require\cancel \cancel{x}+1-\cancel{x}+1=2<\sqrt{4x-1}\left(\sqrt{x+1}+\sqrt{x-1}\right).\tag{$\star$}$$

Once we check $(\star)$ holds for $x=1$ we are done, since clearly its RHS is increasing.

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For $x\ge1$,

$$l(x):=\sqrt{x+1}-\sqrt{x-1}=\dfrac2{\sqrt{x+1}+\sqrt{x-1}}\le\sqrt2$$ and

$$r(x):=\sqrt{4x-1}\ge\sqrt3.$$

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