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Question Number 88067 by A8;15: last updated on 08/Apr/20

Answered by MJS last updated on 08/Apr/20

$$x=\pm \sqrt{a^2+b^2+c^2}$$$$\mathrm{it}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}$$

Commented by john santu last updated on 08/Apr/20

$$supereasysir$$$$$$

Commented by MJS last updated on 08/Apr/20

$$\mathrm{in}\mathrm{this}\mathrm{case},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{but}\mathrm{nearly}\mathrm{impossible}$$$$\mathrm{to}\mathrm{calculate}.{\mathrm{you}}^{\prime }\mathrm{d}\mathrm{have}\mathrm{to}\mathrm{square}3\mathrm{times}$$$$\mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{an}\mathrm{equation}\mathrm{of}{4}^{\mathrm{th}}\mathrm{degree}\mathrm{in}$$$$x^2,{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{I}\mathrm{would}\mathrm{be}\mathrm{able}\mathrm{to}\mathrm{find}\mathrm{the}$$$$\mathrm{root}{x}^2=a^2+b^2+c^2\mathrm{in}\mathrm{all}\mathrm{that}\mathrm{mess}...$$

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