+0  
 
0
16
2
avatar+1199 

Let a and b be the roots of the quadratic 2x^2 - 8x + 7 = x^2 + 15x + 23.  Compute a^4 + b^4.

 Feb 15, 2024
 #1
avatar+144 
+2

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 blush

 Feb 16, 2024
edited by BlackjackEd  Feb 16, 2024
 #2
avatar+128732 
+1

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

 

cool cool cool

 Feb 16, 2024

2 Online Users

avatar
avatar