Find a such that {x1}2+{x2}2 takes the minimal value where x1,x2 are solutions to x2−ax+(a−1)=0.

FiessyFrimatsd0

FiessyFrimatsd0

Answered

2022-01-31

Find a such that {x1}2+{x2}2 takes the minimal value where x1,x2 are solutions to x2ax+(a1)=0.

Answer & Explanation

gekraamdbk

gekraamdbk

Expert

2022-02-01Added 13 answers

(xx1)(xx2)=0
since these are the roots, so:
x2(x1+x2)x+x1x2=0
so x1+x2=a x1x2=a1
now we can say: (x1+x2)2=x12+2x1x2+x22=(x12+x22)+2(a1)=a2
which finally gives us:
y=x12+x22=a22a+2
now to find the minimum differentiate:
dyda=2a2
so for y=0we get a=1 giving min(x12+x22)=1.
coolbananas03ok

coolbananas03ok

Expert

2022-02-02Added 20 answers

From x2ax+(a1), we know that any root satisfies
x2=ax(a1)
x12+x22=(ax1(a1))+(ax2(a1))=a(x1+x2)2a+2.
From x2ax+(a1)=(xx1)(xx2)=x2(x1+x2)x+x1x2,
we have x1+x2=a, so
x12+x22=a(a)2a+2=a22a+11+2=(a1)2+1,
which is a nonnegative term plus 1, so is minimized when the nonnegative term is zero, that is, when a=1.

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