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# Help please!! Quick

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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...

Feb 24, 2018

#1
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Deleted....

Feb 24, 2018
edited by Guest  Feb 24, 2018
edited by ElectricPavlov  Feb 24, 2018
#2
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$$x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3}$$

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

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

Feb 24, 2018
edited by hectictar  Feb 24, 2018
#3
+99586
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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

Feb 24, 2018
#4
+99586
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Sorry, hectictar....I was working on this and then I saw your answer when I submitted mine....at least we agree !!!

Feb 24, 2018
edited by CPhill  Feb 25, 2018
#5
+7354
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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