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Question Number 19643 by Tinkutara last updated on 13/Aug/17
Find the real solution of the equation  (√(17 + 8x − 2x^2 )) + (√(4 + 12x − 3x^2 )) = x^2   − 4x + 13.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\sqrt{\mathrm{17}\:+\:\mathrm{8}{x}\:−\:\mathrm{2}{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{4}\:+\:\mathrm{12}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} }\:=\:{x}^{\mathrm{2}} \\ $$$$−\:\mathrm{4}{x}\:+\:\mathrm{13}. \\ $$
Answered by ajfour last updated on 13/Aug/17
(√u)+(√v)=u−v  ⇒  (√u)−(√v)=1  (√(5^2 −2(2−x)^2 )) =1+(√(4^2 −3(2−x)^2 ))   let (2−x)^2 =t  ⇒25−2t=1+16−3t+2(√(16−3t))  ⇒  (t+8)^2 =64−12t  ⇒  t^2 +28t=0  or     t=0, −28 ;  but t≥0 ,  so       t=0  ,  x=2 .  At x=2 ,  17+8x−2x^2  > 0  and  4+12x−3x^2  > 0  so no objection to  x=2 .
$$\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}=\mathrm{u}−\mathrm{v} \\ $$$$\Rightarrow\:\:\sqrt{\mathrm{u}}−\sqrt{\mathrm{v}}=\mathrm{1} \\ $$$$\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} }\:=\mathrm{1}+\sqrt{\mathrm{4}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{let}\:\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} =\mathrm{t} \\ $$$$\Rightarrow\mathrm{25}−\mathrm{2t}=\mathrm{1}+\mathrm{16}−\mathrm{3t}+\mathrm{2}\sqrt{\mathrm{16}−\mathrm{3t}} \\ $$$$\Rightarrow\:\:\left(\mathrm{t}+\mathrm{8}\right)^{\mathrm{2}} =\mathrm{64}−\mathrm{12t} \\ $$$$\Rightarrow\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{28t}=\mathrm{0} \\ $$$$\mathrm{or}\:\:\:\:\:\mathrm{t}=\mathrm{0},\:−\mathrm{28}\:;\:\:\mathrm{but}\:\mathrm{t}\geqslant\mathrm{0}\:,\:\:\mathrm{so} \\ $$$$\:\:\:\:\:\mathrm{t}=\mathrm{0}\:\:,\:\:\mathrm{x}=\mathrm{2}\:. \\ $$$$\mathrm{At}\:\mathrm{x}=\mathrm{2}\:,\:\:\mathrm{17}+\mathrm{8x}−\mathrm{2x}^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\mathrm{and}\:\:\mathrm{4}+\mathrm{12x}−\mathrm{3x}^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{no}\:\mathrm{objection}\:\mathrm{to}\:\:\mathrm{x}=\mathrm{2}\:. \\ $$
Commented by aknabob last updated on 13/Aug/17
pls its not clear
$${pls}\:{its}\:{not}\:{clear} \\ $$
Commented by Tinkutara last updated on 13/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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