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let alpha, beta be the root of x2+(3-k)x-k=0. The value of k for which (alpha)2+(beta)2 is minimum, is

let alpha, beta be the root of x2+(3-k)x-k=0. The value of k for which (alpha)2+(beta)2 is minimum, is 

Grade:12th pass

1 Answers

Rahul
32 Points
6 years ago
let a and b be roots... (`^` means raised to the power) we have, a+b=-(3-k) and ab =-k(a+b)^2=9+k^2-6ka^2+b^2+2ab=k^2-6k+9a^2+b^2-2k=k^2-6k+9a^2+b^2=k^2-4k+4-4+9a^2+b^2=(k-2)^2+5now a^2+b^2 will be minimum when (k-2)^2 will be zero. therefore, minimum value is 5

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