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# help

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

Aug 25, 2018

#1
+391
+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
+391
+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

dierdurst Aug 25, 2018
#2
+100587
+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

Aug 25, 2018