If a^2 - b^2 = 8 and a*b = 2, find a^4 + b^4.
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calendarEducator since 2008
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Given that:
a^2 - b^2 = 8 ............(1)
ab = 2..................(2)
We need to determine a^4 + b^4
Let us square equation (1).
==> (a^2 - b^2)^2 = 8^2
==> a^4 - 2a^2 b^2 + b^4 = 64
==> a^2 + b^4 = 64 + 2(ab)^2
But from (2) we know that ab= 2
==> a^4 + b^4 = 64 -+2(2^2)
==> a^4 + b^2 = 64 + 8 = 72
==> a^4 + b^4 = 72
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calendarEducator since 2010
write12,544 answers
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We have a^2 - b^2 = 8 and ab = 2
Now (a - b)^2 = a^2 + b^2 + 2ab
=> (a^2 - b^2)^2 = a^4 + b^4 - 2*a^2*b^2
a^4 + b^4
=> (a^2 - b^2)^2 + 2*a^2*b^2
=> 8^2 + 2* (ab)^2
=> 64 + 2*2^2
=> 64 + 8
=> 72
Therefore a^4 + b^4 = 72
Q:If a^2 - b^2 = 8 and a*b = 2, find a^4 + b^4 (edited ).
A:
a^2-b^2 = 8
Therefore (a^2+b^2)^2 = 8^2 = 64.
=> a^4+b^4-2a^2b^2 = 64.
=> a^4+b^4 = 64 + 2a^2b^2
= > a^4+b^4 = 64+ 2(ab)^2
=> a^4+b^4= 64 +2(2)^2, as ab = 2 by data.
=> a^4+b^4 = 64+2*4 = 64 + 8 = 72.
=> a^4+b^4 = 72.
Therefore a^4+b^4 = 72.
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