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.
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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:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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