Solve the equation
$$x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x},$$
where $x \in \mathbb{R}$.
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Are you looking for real or complex solutions? – Bart Michels Sep 04 '13 at 08:15
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What have you done ? What's your ideas for this problem ? – user37238 Sep 04 '13 at 08:17
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I'm assuming we are solving over the reals.
Set $a=\sqrt{3-x}$ and likewise for $b$ and $c$. Then $x=ab+bc+ca$ and we see $$a^2+x = 3,\qquad b^2+x=4,\qquad c^2+x=5$$ Substituting for $x$, we find $$(a+b)(a+c) = 3,\qquad (b+c)(b+a)=4,\qquad (c+a)(c+b)=5$$ Thus $$(a+b)(a+c)(b+c)=\sqrt{60}$$ This leads to $$a+b=\frac{\sqrt{60}}{5},\qquad b+c=\frac{\sqrt{60}}{3},\qquad a+c=\frac{\sqrt{60}}{4}$$ So $$a = a+b+c - (b+c) = \frac{47}{2\sqrt{60}} - \frac{20}{\sqrt{60}} = \frac{7}{2\sqrt{60}}$$ Thus $$x = 3-a^2 = 3 - \frac{7^2}{240} = \boxed{\frac{671}{240}}$$
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1Likewise, with $3,4,5$ replaced by $\alpha,\beta,\gamma$, we obtain $x=\frac{2(\alpha+\beta+\gamma)\alpha\beta\gamma-(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)}{4\alpha\beta\gamma}$ – Bart Michels Jan 26 '17 at 22:27