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Given x_1,x_2 are the real roots of x^2-2mx+(m^2+2m+3)=0, please determine the minimum value of (x_1)^2+(x_2)^2.

 Oct 5, 2018
edited by yasbib555  Dec 9, 2018
 #1
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+2

Let

 

x1  =      [( 2m  + √ [ 4m^2  - 4(1)(m^2 + 2m + 3 ) ]  ) / 2 ]

 

x2  =  [( 2m  - √ [ 4m^2  - 4(1)(m^2 + 2m + 3 ) ]  ) / 2 ]

 

(x1)^2  + (x2)^2  =

 

2 * [4m^2 + 4m^2  - 4m^2 - 8m - 12 ] / 4  =

 

[4m^2 - 8m - 12 ] / 2  =

 

2m^2 - 4m - 6

 

The m value that minimizes this is   4 / (2 * 2)  =  4 / 4   = 1

 

So...the minimum value of  (x1)^2 + (x2)^2  = 2(1)^2 - 4(1) - 6  =  -8

 

 

cool cool cool

 Oct 5, 2018

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