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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
+151
+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

Feb 16, 2024
edited by BlackjackEd  Feb 16, 2024
#2
+129829
+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

Feb 16, 2024