+1133 Let a and b be the roots of the quadratic 2x^2 - 8x + 7 = x^2 + 15x + 23. Compute a^4 + b^4.
+144 Let \(a\) and \(b\) be the roots of the quadratic \(2x^2 - 8x + 7 = x^2 + 15x + 23.\) Compute \(a^4 + b^4\).
We can first calculate the roots of the quadratic.
They are:
\(\frac{23+\sqrt{593}}{2}\) and \(\frac{23-\sqrt{593}}{2}\).
Now we can calculate the \(a^4 + b^4\) part.
\((\frac{23+\sqrt{593}}{2})^4+(\frac{23-\sqrt{593}}{2})^4\)
\(=\frac{314209+12903\sqrt{593}}{2}+\frac{314209-12903\sqrt{593}}{2}\)
\(=\boxed{314209}\)
Our final answer is \(\boxed{314209}.\)
*There is probably a much faster and more efficient way to solve this that someone more educated than I am will be able to teach you clearly 
+128577 Thx, BlackjackEd....here's another way without having to actually calculate the roots
Simplify as
x^2 - 23x - 16 = 0
Product of the roots = ab = -16
2ab = -32
2(a^2b^2) = 2(ab)^2 = 2(-16)^2 = 512
Sum of the roots = a + b = 23
(a + b)^2 = 529
(a^2 + 2ab + b^2) = 529
(a^2 + b^2) + 2ab = 529
(a^2 + b^2) + (-32) = 529
(a^2 + b^2) - 32 = 529
(a^2 + b^2) = 529 + 32 = 561
(a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4
(a^2 + b^2)^2 - 2a^2b^2 = a^4 + b^4
(561)^2 - 512 = a^4 + b^4
314209 = a^4 + b^4
