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The solution of the equation

$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$

is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.

Actually, if we take $f(x)=\sqrt{(x+1)} -\sqrt{(x-1)} -\sqrt{(4x-1)}$ then we can see that $f(x)$ is defined when $x \geq 1$ and $f(1) \geq 0\mbox{ and }f'(x) \geq 0$ so, the function is monotone increasing and it will never appear zero.

so, my question is , In this type of equation where the solution actually does not exist, then why should we get this type of solution?

my solution procedure is, $$ \begin{align} \sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\ \implies 2x-2\sqrt{x^2-1}&=4x-1\\ \implies {-2}\sqrt{x^2-1}&= 2x-1\\ \implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4 \end{align}$$

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As you wrote it, $f(x)$ is monotone decreasing with $f(1) <0$ –  Claude Leibovici Feb 22 at 6:09
    
$f(x)$ is not defined when $x < 1$ –  ROBINSON Feb 22 at 6:12
2  
process your solution procedure in reverse order: substitute your solution for x in the equation. when it first hapens that it is not a solution then figure out what happened –  miracle173 Feb 22 at 6:27
    
In simple words, square the equation only when you know the signs of both sides are the same. Only when both sides satisfy one of $\ge 0$, $\le 0$ together can you square the equation. $(-1)^2=1^2$, e.g., doesn't imply that $-1=1$. Let the equation you've squared be $LHS=RHS$. The case when $LHS=RHS$ in this case is impossible, so squaring, instead of showing you two solutions, one of which is correct and the other one that doesn't satisfy $LHS=RHS$, brings you only one solution. That is, the solution to $-LHS=RHS$. Hence $x=\frac{5}{4}$ satisfies $-LHS=RHS$ only, which is not what you want. –  mathh Feb 22 at 15:30
    
Your assertion that $f(1)\geq 0$ is false. $f(1) = \sqrt{2} - \sqrt{0} - \sqrt{3}$ and $\sqrt{2} -\sqrt{3}<0$. –  Glen_b Feb 22 at 23:54

7 Answers 7

up vote 16 down vote accepted

Let's consider a more simple example, to understand. Given the equation $$ x = 1 $$ you can take the square of both sides: $$ x^2 = 1 $$ and find two solutions: $$ x=1 \qquad x=-1. $$

This happens because the operation $x\mapsto x^2$ is not invertible. If you apply a non invertible function to an equation, the number of solutions might increase.

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$-\frac{x}{x}= \frac{x}{x}$ has no solutions at all, since $-\frac{x}{x}\neq \frac{x}{x}$ no matter what $x$ is.

But we can square both sides, and then what happens?

$\frac{x^2}{x^2}= \frac{x^2}{x^2}$ is an equation that is true for all nonzero numbers.

By applying a non-invertible operation to both sides, we can turn an equation with no solutions into one with uncountably infinitely many solutions.

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Whenever we square, we immediately introduce extraneous root

Observe that $\displaystyle\frac54$ is actually a root of $$\sqrt{x+1}=\sqrt{4x-1}-\sqrt{x-1}$$

Also, observe that $\displaystyle2x-1=-2\sqrt{x^2-1}\le0\implies 2x\le1\iff x\le\frac12$ for real $x$

But, $\displaystyle{\sqrt{x-1}}$ is not real unless $x\ge1$

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The process of solving an equation is basically that of inversion: you successively apply (inverse) functions to both sides of the equation until you reach a point where the solution is clear. This process depends on each successive equation (upon applying various inverses successively) being equivalent to the previous one so that the final equation $x=\ldots$ is equivalent to the original equation. However, when you apply non-invertible operations (such as $x\mapsto x^{2}$, i.e. squaring both sides), you don't get an equivalence between the equation before squaring and the equation after squaring: you get a forward implication, which is to say that the final equation $x=\ldots$ does not imply the previous equation(s) prior to squaring, it is only implied itself by the previous chain of equations.

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Write the equation as $$ \sqrt{x+1}=\sqrt{x-1}+\sqrt{4x-1} $$ Then you must have \begin{cases} x+1\ge0\\ x-1\ge0\\ 4x-1\ge0 \end{cases} which boils down to $x\ge1$. Now square, you're sure not to add spurious solutions, because both sides represent non negative numbers: $$ x+1=x-1+4x-1+2\sqrt{(x-1)(4x-1)} $$ or $$ -4x+3=2\sqrt{(x-1)(4x-1)} $$ Now the right hand side is non negative, so also the left hand side must be, which means $$ -4x+3\ge0 $$ or $x\le 3/4$. With the previous limitation, this has the consequence that no solution can exist.

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An equation $F(x)=0$ is an implicit way of defining a set $S$. What we want is an explicit presentation of $S$, say, a list $S=\{a,b,c\}$.

Algebraic manipulations of the form $$F(x)=0\quad\Rightarrow \quad G(x)=0\quad\Rightarrow\quad\ldots\quad\Rightarrow \quad x\in S'\ ,\tag{1}$$ where $S'$ is a certain finite list (or similar object) do not prove that $S=S'$. They only prove that each $x\in S$ also is in $S'$, in other words: Such manipulations only prove $S\subset S'$. When $S'$ is a finite set it is usually simple to decide which $x\in S'$ are actually solutions of the original problem.

When the arrows in $(1)$ are reversible, i.e. can be replaced by $\Leftrightarrow$'s with no harm done, then we of course have $S'=S$, and no extra verification is necessary.

Another instance where $S'$ is automatically the correct solution set is the following: The given equation $F(x)=0$ is of a type for which we have a general theory guaranteeing "exactly two solutions" (as in the case of a quadratic equation) or "exactly one solution", or "a solution space of dimension $d$". When the $S'$ found in $(1)$ satisfies the requirements promised by the general theory then automatically $S'=S$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Indeed, you are solving $\large 4!!!$ equations due to the 'squaring' in your procedure: $$ \root{x + 1} \pm \root{x - 1} = \pm\root{4x - 1} $$ For example, $x = 5/4$ satisfies $\root{x + 1} + \root{x - 1} = \root{4x - 1}$ since $$ \root{{5 \over 4} + 1} \color{#f00}{\Large +} \root{{5 \over 4} - 1} = {3 \over 2} + \half= 2= \color{#f00}{\Large +}\root{4\times{5 \over 4} - 1} $$

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$10 \uparrow (10 \uparrow (10 \uparrow 1.4))$ equations –  Hugh Sep 16 at 5:44

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