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a^2 - b^2 = 8 and a*b = 2, find a^4 + b^4.

 Aug 25, 2018

Best Answer 

 #1
avatar+374 
+2

We have \({a}^{2} - {b}^{2} = 8\), which we can square both sides to get \(a^4 + b^4 - 2a^2b^2 = 8^2\). Since \(ab = 2\), we can square both sides again to get \(a^2b^2 = 2^2\). Then, we can subsitute that value into the equation; \(a^4 + b^4 - 2(4) = 8^2\) -> \(a^4 + b^4 - 8 = 64\) -> \(a^4 + b^4 = 72\).

 

- Daisy

 Aug 25, 2018
 #1
avatar+374 
+2
Best Answer

We have \({a}^{2} - {b}^{2} = 8\), which we can square both sides to get \(a^4 + b^4 - 2a^2b^2 = 8^2\). Since \(ab = 2\), we can square both sides again to get \(a^2b^2 = 2^2\). Then, we can subsitute that value into the equation; \(a^4 + b^4 - 2(4) = 8^2\) -> \(a^4 + b^4 - 8 = 64\) -> \(a^4 + b^4 = 72\).

 

- Daisy

dierdurst Aug 25, 2018
 #2
avatar+98091 
+2

a^2  - b^2   = 8   and  ab  = 2 

 

So 

 

(a^2 - b^2) (a^2 - b^2)  =  (8) * (8)

 

a^4  - 2(ab)^2 + b^4  = 64

 

a^4 + b^4  = 64 + 2(ab)^2

 

a^4 + b^4  =  64 + 2(2)^2

 

a^4 + b^4  = 64 + 8

 

a^4 + b^4  = 72

 

 

cool cool cool

 Aug 25, 2018

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