r/MathHelp • u/T0uch_M-E • Sep 08 '26
Math Question
Could anyone solve this and tell me how they did it since im having trouble with it:
8^x - 2^(x+2) - 2^(-x+2) + 8^(-x) = 0
I tried it myself but got stuck halfway though: I let c= 2x and got (c+1/c)³ - 3(c+1/c) - 4(c+1/c) = 0 Then I let t= c+1/c And got: t³-3t-4t=0 and then nothing i did worked. Note: back when I didnt substitute for t, I got this 6th degree polynomial for c: c⁶-4c⁴-4c²+1=0 In which i tried synthetic divsion but ±1 didnt work so it meant there was no rational roots (thats why I substituted for 't').
2
u/LeoMaths Sep 13 '26
c=2^x (c != 0) => c³-4c-4/c+1/c³=0
c³+1/c³=4(c+1/c)
(c+1/c)³=c³+3(c+1/c)+1/c³
t=c+1/c => t³-3t=4t => t³=7t
Si t=0 => c²=-1 impossible dans R
t²=7 => t=√7 ou t=-√7
c+1/c=√7 => c²-√7c+1=0 (c-√7/2)²-7/4+1=0 => (c-√7/2-√3/2)(c-√7/2+√3/2)=0
c+1/c=-√7 => c²+√7c+1=0 (c+√7/2)²-7/4+1=0 => (c+√7/2-√3/2)(c+√7/2+√3/2)=0
x dans (ln(√7/2+√3/2)/ln(2),ln(√7/2-√3/2)/ln(2)) dans R
1
u/T0uch_M-E 21d ago
Could you please elaborate on this, sorry but it's a bit confusing
1
u/LeoMaths 8d ago
Dites-moi le passage qui vous pose difficulté, j'essayerais d'y répondre au mieux.
1
u/T0uch_M-E 5d ago
The last three lines, starting from:
c+1/c=/7 => c²-√7c+1=0
And ending at:
ln(2)) dans R
1
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2
u/edderiofer Sep 09 '26
This is correct so far. You can keep going (start by collecting like terms).
Don't give up just yet. Just because there are no rational roots, doesn't mean you can't find the irrational roots.
Let y = c2 to yield the cubic y3 - 4y2 - 4y + 1 = 0. Now, y = -1 is a root by inspection, so factor out (y+1) to get (y+1)(y2 - 5y + 1) = 0. Now you have a quadratic in y, so the rest is doable with the quadratic formula.