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Excuse me sir/ mamThis que to fin minimum value...Plz may u help meh

Harsh Jain , 7 Years ago
Grade 11
anser 1 Answers
Sujit Kumar

Last Activity: 7 Years ago

x^2 - y^2=6-2xy
(x^2+y^2)^2=(x^2-y^2)^2 + 4x^2y^2
Substituting \ the \ value \ of \ x^2 - y^2
 
(x^2+y^2)^2=4(9+2x^2y^2-6xy)
Let \ us \ consider \ xy=a
=>(x^2+y^2)^2=4(2a^2-6a+9)
 
We \ know \ the \ minimum \ value \ of \ an \ equation = \frac{-D}{4a}
Where \ D=b^2-4ac
=>min. \ value \ of \ 2a^2-6a+9 \ is \ \frac{9}{2}
Therefore \ (x^2+y^2)^2= 4(9/2)=18
Ans: \ The \ minimum \ value \ of \ (x^2+y^2)^2 \ is \ 18
 

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