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Question Number 84820 by Power last updated on 16/Mar/20

Answered by MJS last updated on 16/Mar/20

super easy (√(2x−5))=t ⇔ x=(1/2)t^2 +(5/2)∧t≥0 (√((1/2)t^2 +(5/2)−2+t))+(√((1/2)t^2 +(5/2)+2+3t))=7(√2) (√((1/2)t^2 +t+(1/2)))+(√((1/2)t^2 +3t+(9/2)))=7(√2) (√(((t+1)^2 )/2))+(√(((t+3)^2 )/2))=7(√2) ((t+1)/(√2))+((t+3)/(√2))=7(√2) 2t+4=14 t=5 x=15

supereasy2x−5=t⇔x=12t2+52∧t⩾012t2+52−2+t+12t2+52+2+3t=7212t2+t+12+12t2+3t+92=72(t+1)22+(t+3)22=72t+12+t+32=722t+4=14t=5x=15

Commented by Power last updated on 16/Mar/20

thanks

thanks

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