MHB Equation 2: Prove that ##x^2+2x\sqrt x+3x+2\sqrt x+1=0##

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The equation x^2 + 2x√x + 3x + 2√x + 1 = 0 has been checked with WolframAlpha, which indicates there are no real solutions. Participants suggest proving this by either using Ferrari's method or graphing the equation. One contributor notes that if √x is not purely imaginary, then x must be complex. The discussion highlights the challenges in finding real solutions to the equation. Ultimately, the consensus leans towards graphing as a preferred method for analysis.
solakis1
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I checked the following equation with Wolfram\Alpha and the answer was no real solution
How can we prove that?
$x^2+2x\sqrt x+3x+2\sqrt x+1=0$
 
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solakis said:
I checked the following equation with Wolfram\Alpha and the answer was no real solution
How can we prove that?
$x^2+2x\sqrt x+3x+2\sqrt x+1=0$
You pretty much are stuck with two choices: Ferrari's method or graphing. I'd choose graphing!

-Dan
 
$f(x) = x^2 + 2x\sqrt x + 3x + 2\sqrt x + 1 = (x + \sqrt x + 1)^2$, so if $f(x) = 0$ then $x + \sqrt x + 1 = 0$. That is a quadratic equation for $\sqrt x$, with solutions $\sqrt x = \frac12(-1 \pm i\sqrt3)$. If $\sqrt x$ is non-real then so is $x$. Therefore the equation $f(x) = 0$ has no real solutions.
 
[sp]Well, if $\sqrt{x}$ is not pure imaginary then x is complex, anyway.[/sp]

Nice catch!

-Dan
 
ha, ha so easy solution:mad:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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