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In solving $\sqrt{x^2 + 4x - 3} = 1 - 2x$ :

To eliminate the square root, we square both sides, and this yields $x^2 + 4x - 3 = (1 - 2x)^2 = 1 - 4x + 4x^2 <=> -3x^2 + 8x -4 = 0$, which has solution set $S =$ {2/3 , 2}, but, returning to the original equation: any of the two makes $1 - 2x$ negative, and therefore the original equation which ought to be equivalent to the one we solved is not solved with the same solution set. Where is the error?

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  • $\begingroup$ -3 is a square root of 9, and 1/3 is a square root of 1/9, so your statement that they don't solve the original equation is not true. $\endgroup$ – Nij Sep 13 '16 at 20:33
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    $\begingroup$ The square function is not injective. Hence you may gain extraneous solutions. $\endgroup$ – MathematicsStudent1122 Sep 13 '16 at 20:36
  • $\begingroup$ When you square both sides you are allowing for the extraneous and wrong solutions to $-\sqrt {x^2+4x-3} =1-2x$. These are twice as many solutions and you have avoid these solutions by restricting 1-2x >= 0 so x <= 1/2. There are no solutions. 2/3 and 2 are solutions to $\sqrt{stuff} = 2x-1$, a different equation altogether. $\endgroup$ – fleablood Sep 13 '16 at 21:10
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By definition $\sqrt{X}$ is the non-negative $Y$ such that $Y^2=X$ so the solution of $\sqrt{x^2 + 4x - 3} = 1 - 2x$ require that $1-2x\ge 0$. You have found that this cannot be done for the solution of the squared equation , so the given equation has no solutions.

If you have some doubt about this result think to your equation as the intersections of a half-hyperbola (of positive $y$) of equation $y=\sqrt{x^2 + 4x - 3}$ and a straight line of equation $y=1 - 2x$. It is well possible that the line has no intersection with the half hyperbola, also if it has intersection withe the other half hyperbola.

Seethe figure:

enter image description here

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    $\begingroup$ $\sqrt{X}$ actually is defined to be non-negative, for the nitpicking. Otherwise $\sqrt0$ wouldn't exist. $\endgroup$ – Michael Hoppe Sep 13 '16 at 20:47
  • $\begingroup$ @MichaelHoppe: OK! A little abuse that i edit:) And I add also something. $\endgroup$ – Emilio Novati Sep 13 '16 at 20:57
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You applied a function, say $f$, to both sides of your equality $a=b$ and you claim that $$ f(a)=f(b)\Rightarrow a=b $$ What kinds of functions $f$ satisfy this?

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