<s>[INST] How can I prove this equation has no real solution? I have an equation
$$x^2-5 x+10+(x-4) \sqrt{1+x}=0. \tag{1}$$
Now I am trying to prove this equation has no real solution. 
I tried. Put $t = \sqrt{1+x}$, then, I got
$$t^4+t^3-7 t^2-5 t+16=0. \tag{2}$$
But I can't prove two equations (1) and (2) have not solution. 
How can I prove?
 [/INST] $$x^2-5x+10-(4-x)\sqrt{x+1}$$ 
now note: for roots to exist $-1<x<4$ thus $(4-x)>0$ and $\sqrt{x+1}>0$ now using our favourite AM-GM
$$x^2-5x+10-(4-x)\sqrt{x+1}>x^2-5x+10-\frac{(4-x)^2+(x+1)}{2}=\frac{2x^2-10x+20-16-x^2+8x-x-1}{2}=\frac{x^2-3x+3}{2}$$ now discriminant of the quadratic is negative and first coefficient is positive thus no roots exist and quadratic is greater than zero thus your function has no roots
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