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Suppose r and s are the values of x that satisfy the equation x^2 - 2mx + (m^2 + 2m + 3) = 0
for some real number m. Find the minimum real value of r^2 + s^2.

i thought 4 was the answer but it is incorrect...

Guest Feb 24, 2018
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5+0 Answers

 #1
avatar+12266 
0

Deleted....

ElectricPavlov  Feb 24, 2018
edited by Guest  Feb 24, 2018
edited by ElectricPavlov  Feb 24, 2018
 #2
avatar+7056 
+2

\(x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3} \)

 

 

\(r = -m + \sqrt{-2m-3} \\~\\ r^2=(-m + \sqrt{-2m-3})(-m + \sqrt{-2m-3}) \\~\\ r^2=m^2-2m\sqrt{-2m-3}+(-2m-3)\)

 

and

 

\(s = -m-\sqrt{-2m-3} \\~\\ s^2=(-m-\sqrt{-2m-3})(-m-\sqrt{-2m-3}) \\~\\ s^2=m^2+2m\sqrt{-2m-3}+(-2m-3)\)

 

 

\(r^2+s^2=m^2-2m\sqrt{-2m-3}+(-2m-3)+m^2+2m\sqrt{-2m-3}+(-2m-3) \\~\\ r^2+s^2=2m^2+2(-2m-3) \\~\\ r^2+s^2=2m^2-4m-6\)

 

The minimum value of  r^2 + s^2  occurs when  m   =   4/( 2 * 2 )   =   1

 

When m = 1 ,   r2 + s2   =   2 - 4 - 6   =   -8

hectictar  Feb 24, 2018
edited by hectictar  Feb 24, 2018
 #3
avatar+86649 
+2

Here's my take on this....

 

The sum of the roots is  r + s  = 2m   .....so....

(r + s)^2  = 

r^2 + 2rs + s^2   = 4m^2    (1)

 

The product of the roots is  r * s   =  m^2 + 2m + 3   (2)  

 

Manipulating (1), we have that

r^2 + s^2  =  4m^2 - 2rs

Sub (2)  into this

r^2 + s^2  = 

4m^2  - 2(m^2 + 2m + 3)   =  

4m^2 - 2m^2 - 4m - 6

2m^2 - 4m - 6         (3)

 

(3)  is minimized when  m = 

 

4/ (2*2)   =  4 / 4   =  1

 

And the minimum value is  2(1)^2 - 4(1) - 6  =  2 - 4 - 6  =  -8  

 

So......

 

The minimum  of   r^2 + s^2   = [ min ] 2m^2 - 4m - 6  =  - 8    

 

 

cool cool cool

CPhill  Feb 24, 2018
 #4
avatar+86649 
+1

Sorry, hectictar....I was working on this and then I saw your answer when I submitted mine....at least we agree !!!

 

 

 

cool cool cool

CPhill  Feb 24, 2018
edited by CPhill  Feb 25, 2018
 #5
avatar+7056 
+1

Haha that's because I was logged out when I started to answer! I'm just glad you got the same thing!

hectictar  Feb 24, 2018

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