+0  
 
0
24
1
avatar+209 

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.

yasbib555  Oct 5, 2018
 #1
avatar+89874 
+1

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

CPhill  Oct 5, 2018

24 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.