(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2+2ab+2b^2 =The answer
(a + b)^2 = a^2 + 2ab + b^2 => square of sums
(a - b)^2 = a^2 - 2ab + b^2 => square of deference
and of course one of most important ones:
a^2 - b^2 = (a - b)(a + b) => difference of squares
Best Answer: (a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2 + 2ab + 2b^2
a^4 + 4b^4 => i.e. 4a^2b^2 ,
a^4 + 4a^2b^2 + 4b^4 => a^2 + 2ab + b^2 = (a + b)^2, if : a = a^2 , b = 2b^2:
(a^2 + 2b^2)^2 = a^4 + 4a^2b^2 + 4b^4 => We can't add or subtract the value to the expression.
a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 =>
(a^2 + 2b^2)^2 - 4a^2b^2 =>
(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) =>
(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)
Greetings!
Given
we have
Squaring both sides, we have
And finally
Note that, when we square both sides, we have to assume that
because we're assuming that this fraction equals a square root, which is positive.
So, if that fraction is positive you'll actually have roots: choose
and you'll have
Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose
and you'll have
Squaring both sides (and here's the mistake!!) you'd have
which is not a solution for the equation, if we plug it in we have
Which is clearly false
Answer:
probably an update which sucks tell me about it lol
