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Let u and v be the solutions to 3x^2 + 5x + 7 = 0.  Find u^2 + v^2.

 Jul 13, 2021
 #1
avatar+18 
+1

 

First, you need to find the solutions.

 

3x2+5x+7=0

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-5 \pm \sqrt{5^2-4*3*7} \over 2*3}\)

x = -(5-i√59)/6, and -(5+i√59)/6

We can rewrite as:

v = -(5+i√59)/6

 u = -(5-i√59)/6

u2+v2 = (-(5-i√59)/6)2+ (-(5+i√59)/6)2

= -17/9

 

SWest

 Jul 13, 2021
 #2
avatar+121065 
+1

Thx, SWest    !!!

 

Here's  another way

 

By Vieta's theorem  :

 

sum of  the  solutions =    -5/3  =  u + v

Square  both sides    25/9   = u^2  + 2uv + v^2      (1)

 

product of the  solutions  =  7/3 =  uv

So   2uv   =   14/3   

Sub this into (1)  and we  have

 

25/9  = u^2  + (14/3)  +  v^2

 

25/9  - 14/3   = u^2  + v^2

 

25/9   -  42/9   =  u^2  +  v^2

 

-17 / 9   = u^2  + v^2

 

cool cool cool

 Jul 13, 2021

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