+0  
 
-1
110
4
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The quadratic 2x^2-3x+27 has two imaginary roots. What is the sum of the squares of these roots? Express your answer as a decimal rounded to the nearest hundredth.

 

Please Help Quick!

 Dec 16, 2020
 #1
avatar+112826 
0

Why do you need help so quickly?

Are you doing a test?

Have you left your homework till the last minute and need someone else to do it for you?

 Dec 16, 2020
 #2
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-1

No, I'm doing an AoPs question, and I have tried to do it myself but I just don't know how to find the roots. Also, this work is due in January, and I am just bored and wondering if I can learn new things by doing a bit of work ahead of time. If you could help me with that it would be much appreciated. Thank you!

 Dec 16, 2020
 #3
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0

Use the Quadratic formula to find the roots

a=2       b =-3      c = 27  

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)       then    x = 3/4 +- 3/4 * i sqrt23  

                                           

         squaring them results in    9/16 +18/16 sqrt23 i - 9/ 16 * 23       and    9/16 - 18/16 sqrt23 i - 9/16*23

                                            then  adding them =  - 198/8 = -99/4 

 Dec 17, 2020
 #4
avatar+117546 
+1

Call the  roots  m and  n

 

By Vieta's Theorem......

 

ax^2  + bc + c

 

Sum  of roots   = m + n  =  -b/a  =   3/2       square both sides   m^2 + 2mn + n^2  = 9/4   (1)

Product of  roots  = mn  =  c/a  =  27/2      and  2mn =  27    (2)

 

So......subbing   (2) into ( 1)   we have

 

m^2  +  27  + n^2  =  9/4

 

m^2 + n^2   =   9/4 - 27

 

m^2  + n^2   =  [9 - 108 ]  / 4   =   - 99 / 4

 

 

cool cool cool

 Dec 17, 2020

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