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x = 1, just treat as a partial factorial, eazy peazy ... Part of mathematics is recognizing relations to simplify expressions. You made something very simple into an extremely complex conglomeration of algebraic calculations.
You are showing very lengthy method my dear. The short method is take product of first and last bracket together and middle bracket together. Put x^2 +5x = m and you will get equation m^2 +10m -- 96 =0. Now solve this equation.
I’m not sure he understood your question. But most likely it was a typing error in the very first line(missing to write the letter 2(two), and that might be because of the cognitive similarity of the sqrt and the letter 2(if you rotate the digit 2 about 45 degrees counterclockwise ), as both the thumbnail, and all further calculations rely on sqrt(28) instead of sqrt(8).
Good work! But I think that follow solution is easy Eq1: 222^x + 222^(2x) = 222^(3x) let y=222^x then 222^x=y 222^(2x)=y^2 222^(3x)=y^3 so Eq1: y + y^2 = y^3 y^3 - y^2 - y = 0 y*(y^2 - y - 1) = 0 then y=0 or y^2 - y - 1 = 0 y^2 - y - 1 = 0 Delta = 1 -4*(-1) = 5 y = (1 + - sqr(5) ) /2 1) y=0 222^x=0 not exit x (or x = - infinity) 2) y = (1 + sqr(5) ) /2 222^x= (1 + sqr(5) ) /2 log (222^x) = log((1 + sqr(5) ) /2) x*log(222) = log((1 + sqr(5) ) /2) x= log((1 + sqr(5) ) /2) / (log(222)) or x = log log((1 + sqr(5) ) /2) 222 3) y = (1 - sqr(5) ) /2, but in this case y < 0 222^x= (1 + sqr(5) ) /2 log (222^x) = log((1 - sqr(5) ) /2) but no real solution because log of negative number is imaginary! x= log((1 - sqr(5) ) /2) / (log(222)) is imaginary solution! THEN the only real solution is x = log (1 + sqr(5) ) /2 222
JUST SEE THAT THE MAXIMUM POWER OF 2 IS 7 HERE THAT WOULD GIVE 128.SUTRACT FROM 148 TO GET 2O,WHICH IS 16 PLUS 4.THUS YOU GET 2 TO THE POWER 4 AND 2 TO THE POWER 2. MARK A B C AS 7 4 AND 2 NO RIGOROUS CALCULATIONS 🤩😄😀😆😅
Good solution! Another solution: (x+2)^4+(x+1)^4=17 let y=x+1 (y+1)^4+y^4=17 easy to see that y=1 is solution ==> (2^4)+1^4=16+1=17 but: (y+1)^4=y^4+4y^3+6y^2+4y+1 arrange then equation y^4+4y^3+6y^2+4y+1 + y^4=17 2y^4+4y^3+6y^2+4y-16=0 y=1 is root, using the synthetic division: 2 4 6 4 -16 | 1 2 6 12 16 | -------------------------| 2 6 12 16 | 0 | equation now: (y-1)*(2*y^3+6y^2+12y+16) = 0 but 2*y^3+6y^2+12y+16=0 (divide by 2) y^3+3y^2+6y+8=0 if exist a integer root this root is divisor of 8, and root need to be negative because all coeficients are positive check -1: is not root, because f(-1)=-1+3-6+8=+4 but -2 is root, because f(-2)=-8+12-12+8=0 y=-2 is root, using the synthetic division 1 3 6 8 | -2 -2 -2 -8 | ---------------| 1 1 4 | 0 | equation: (y+2)*(y^2+y+4)=0 but y^2+y+4=0 Delta=1^2-4*1*4=1-16=-15 and y=(-1+-sqr(15)i)/2 then y=1; y=-2 and y=(-1+-sqr(15)i)/2 are the solutions but y=x+1 then x=y-1 so x=0; x=-3; x=(-3+-sqr(15)i)/2 are the solutions
