Solving for a and b: Fast Method for a^4 + b^4 Calculation

AI Thread Summary
To find the value of a^4 + b^4 given a-b = 5 and ab = 2, a faster method involves using the identity a^4 + b^4 = (a-b)^4 + 4ab(a^2 + b^2) - 6(ab)^2. By substituting the known values, (a-b)^4 becomes 625 and 4ab(a^2 + b^2) can be simplified using the relationship between a and b. The calculation ultimately leads to a straightforward expression that avoids solving for a and b directly. This method efficiently computes a^4 + b^4 without requiring the individual values of a and b.
Taturana
Messages
108
Reaction score
0

Homework Statement



Given that a-b = 5 and ab = 2, what is the value of a^4 + b^4?

Homework Equations



The Attempt at a Solution



Doing the math I know that
a_{1} = \frac{5+\sqrt{33}}{2}
a_{2} = \frac{5-\sqrt{33}}{2}

b_{1} = \frac{-5+\sqrt{33}}{2}
b_{2} = \frac{-5-\sqrt{33}}{2}

So my question is: there is any fast way to do a^{4} + b^{4}?
 
Physics news on Phys.org
You can expand a^4+b^4 so you never have to solve for a and b, as follows:
a^4+b^4=(a-b)^4+4a^3b-6a^2b^2+4ab^3=(a-b)^4+4ab(a^2+b^2)-6a^2b^2=(a-b)^4-6(ab)^2+4ab((a-b)^2+2ab) = (a-b)^4+2(ab)^2+4ab(a-b)^2

Now you just plug in what you know for (a-b) and ab.
 
Back
Top