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Question Number 182593 by srikanth2684 last updated on 11/Dec/22

Commented by srikanth2684 last updated on 14/Dec/22

sol pls

$${sol}\:{pls} \\ $$

Answered by TheSupreme last updated on 15/Dec/22

(√)(x−2)^2 +6^2 +(√((12−x)^2 +4^2 ))=d (x−2)^2 +6^2 −(12−x)^2 −4^2 −d^2 =−2d(√((12−x)^2 +4^2 )) −4x+4+36−12^2 +24x−16−d^2 =−2d(√((12−x)^2 +4^2 )) 20x−144+40−16−d^2 =−2d(√((12−x)^2 +4^2 ))

$$\sqrt{}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\sqrt{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }={d} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\left(\mathrm{12}−{x}\right)^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} −{d}^{\mathrm{2}} =−\mathrm{2}{d}\sqrt{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} } \\ $$$$−\mathrm{4}{x}+\mathrm{4}+\mathrm{36}−\mathrm{12}^{\mathrm{2}} +\mathrm{24}{x}−\mathrm{16}−{d}^{\mathrm{2}} =−\mathrm{2}{d}\sqrt{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} } \\ $$$$\mathrm{20}{x}−\mathrm{144}+\mathrm{40}−\mathrm{16}−{d}^{\mathrm{2}} =−\mathrm{2}{d}\sqrt{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} } \\ $$

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