Equation

**james_bond** · November 25th 2008, 06:48 AM

Is there any real $x$ that solves $\sqrt{x^4+x+2}=\sqrt[3]{x^5+8x+1}$ ?

(I checked with Wolfram Mathematica and got 2 real (and 2 complex) roots.)

Of course I only have to show that there are real root(s) and don't have to find them.

**skeeter** · November 25th 2008, 07:51 AM

$f(x) = \sqrt{x^4 + x + 2} - \sqrt[3]{x^5 + 8x + 1}$

consider the signs of $f(0)$ and $f(1)$ .

**masters** · November 25th 2008, 10:14 AM

Originally Posted by skeeter

$f(x) = \sqrt{x^4 + x + 2} - \sqrt[3]{x^5 + 8x + 1}$

consider the signs of $f(0)$ and $f(1)$ .

And also the sign change from $f(1)$ to $f(2)$

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