SOLUTION: if a,b are the roots of the equation 2x2+6x-10=0.Find a4+b4,a3-b3,a6+b6,a6-b6

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Question 241920: if a,b are the roots of the equation 2x2+6x-10=0.Find a4+b4,a3-b3,a6+b6,a6-b6
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
if a,b are the roots of the equation 2x%5E2%2B6x-10=0.Find a%5E4%2Bb%5E4, a%5E3-b%5E3, a%5E6%2Bb%5E6, a%5E6-b%5E6

Since a, b are roots of 

2x%5E2%2B6x-10=0

they are also roots of the monic polynomial equation found
by dividing through by 2

x%5E2%2B3x-5=0

The sum of the roots is the negartive of the coefficient of x,
-3.  The product of the roots is the constant number or -5
The absolute value of the difference of the roots is 
sqrt%28discriminant%29=sqrt%283%5E2-4%28-5%29%29=sqrt%289%2B20%29=sqrt%2829%29

If we let a be the larger root, then a-b+=+sqrt%2829%29  

So a%2Bb=-3, ab=-5, a-b=sqrt%2829%29

So 

So a%5E2-10%2Bb%5E2=9

So a%5E2%2Bb%5E2=19

Square both sides:

a%5E4%2B2a%5E2b%5E2%2Bb%5E4=361
a%5E4%2Bb%5E4=361-2a%5E2b%5E2
a%5E4%2Bb%5E4=361-2%28ab%29%5E2
a%5E4%2Bb%5E4=361-2%28-5%29%5E2
a%5E4%2Bb%5E4=361-2%2825%29
a%5E4%2Bb%5E4=361-50
a%5E4%2Bb%5E4=311

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But also %28a%2Bb%29%5E3=%28-3%29%5E3=-27

So a%5E3%2Bb%5E3%2B45=-27 or a%5E3%2Bb%5E3=-72




But also %28a%5E3%2Bb%5E3%29%5E2=%28-72%29%5E2=5184,

So a%5E6%2Bb%5E6-250=5184 or

a%5E6%2Bb%5E6=5434

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Edwin