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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?

minhthien_2016
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  • Use Cardan's Formula to solve it. Then show that none of the real roots (if any) is positive. See https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots –  May 20 '16 at 06:10
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    Why the downvote? This is a good question. – Crostul May 20 '16 at 06:13
  • Long (but systematic) method: study the variations of $f\colon\mathbb{R}\to\mathbb{R}$ defined by $$ f(t)=t^4+t^3-7t^2-5t+16 $$ and show that it never cancels (you may have to dfferentiate twice). – Clement C. May 20 '16 at 06:13

1 Answers1

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$$x^2-5x+10-(4-x)\sqrt{x+1}$$ 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

avz2611
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