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Question Number 168946 by bagjagugum123 last updated on 22/Apr/22

Commented by bagjagugum123 last updated on 22/Apr/22

Prove that : w=(√((x+y+z)^2 −4xy))

Provethat:w=(x+y+z)2−4xy

Answered by infinityaction last updated on 22/Apr/22

Commented by bagjagugum123 last updated on 22/Apr/22

Very nice, thank you Sir

Verynice,thankyouSir

Answered by som(math1967) last updated on 22/Apr/22

If figure is square let side =l then z=(1/2)(((l^2 −2x)/l))(((l^2 −2y)/l)) 2l^2 z=l^4 −2l^2 (x+y)+4xy ⇒l^4 −2l^2 (x+y+z)+4xy=0 l^2 =((2(x+y+z)+(√(4(x+y+z)^2 −16xy)))/2) l^2 =(x+y+z)+(√((x+y+z)^2 −4xy)) x+y+z+w =(x+y+z)+(√((x+y+z)^2 −4xy)) ∴w=(√((x+y+z)^2 −4xy))

Iffigureissquareletside=lthenz=12(l2−2xl)(l2−2yl)2l2z=l4−2l2(x+y)+4xy⇒l4−2l2(x+y+z)+4xy=0l2=2(x+y+z)+4(x+y+z)2−16xy2l2=(x+y+z)+(x+y+z)2−4xyx+y+z+w=(x+y+z)+(x+y+z)2−4xy∴w=(x+y+z)2−4xy

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