Can YOU Crack This Rational Equation Challenge? | Algebra

[@Quest3669]

X=( -7+-√13)/2; (-9+-√13)/2

[@gregevgeni1864]

x =-4 +(1±√13)/2 , x =-4 +(-1±√13)/2.
Obviously x ≠ -4 .
The given equation is equivalent to
{[(x+4)+2]•[2(x+4)-3]/(x+4)}⁴ +
+ {[(x+4)-2]•[2(x+4)+3]/(x+4)}⁴ = 82 (1).
Let t = x+4 (*) and the (1) write as
[(t+2)(2t-3)/t]⁴ +[(t-2)(2t+3)/t]⁴=82 [(2t²+t-6)/t]⁴ +[(2t²-t-6)/t]⁴=82
[2(t-3/t)+1]⁴ +[2(t-3/t)-1]⁴=82 (2).
Let u = 2(t-3/t) (**)and the (2) write as
(u+1)⁴ +(u-1)⁴= 82 u⁴+6u²-40=0
(u²+10)(u²-4)=0 => u =±2 .
From (**) =>
case one
2(t-3/t)=2 => t-3/t=1 =>t²-t-3=0=>
t =(1±√13)/2 and x = (1±√13)/2 -4 due to (*)
case two
2(t-3/t)=-2 => t-3/t=-1 => t²+t-3=0 => t =(-1±√13)/2 and x=(-1±√13)/2 -4 due to (*) .

[@Chacal0152]

x=a-4, a ≠ 0, [(a+2)(2a-3)]⁴+[(a-2)(2a+3)]⁴ = 82a⁴ = [2a²-6+a]⁴+[2a²-6-a]⁴ , b =2a²-6 = ka =>
(k+1)⁴+(k-1)⁴ = 82 = 2(k⁴+6k²+1) => k⁴+6k²-40 = 0 = (k²+10)(k²-4) => k² = 4 => k = ±2 =>
2a²±2a -6 = 0 = 2(a²±a-3) => a =½(±1±V13) => x = ½(±1±V13)-4 = ½(-8±1±V13)

[@潘博宇-k4l]

X1=[(13)^(1/2)-7]/2, X2=[-(13)^(1/2)-7]/2, X3=[(13)^(1/2)-9]/2, X4=[-(13)^(1/2)-9]/2.

[@ManojkantSamal]

^= read as to the power
*=read as square root
Let R={(x+6)(2x+5)}/(x/4)
A={(x+2)(2x+11)}/(x+4)
As per question
R^4+A^4=82.....eqn1
Now explain R
{(X+6)(2x+5)}/(x+4)
={2x^2+5x+12x+30}/(x+4)
=(2x^2+17x+30)/(x+4)
Explain A
{(X+2)(2x+11)}/(x+4)
=(2x^2+11x+4x+22)/(x+4)
=(2x^2+15x+22)/(x+4)
Now R-A
={2x^2+17x+30-2x^2-15x-22}/(x-4)
=(2x+8)/(x+4)
=2(x+4)/(x+4)
=2
So,
R-A=2...eqn2
We know that
(R-A)^4=R^4+A^4+(6R^2A^2)-4RA(R^2+A^2)
SO,
2^4=82+(6R^2A^2)-4RA{(R-A)^2+2RA}
SO,
16=82+(6R^2A^2)-4RA(4-2RA)
16-82=(6R^2A^2)-16RA-(8R^2A^2)
-66=(-2R^2A^2)-16RA
-66=-{(2R^2A^2)+16RA}
Let RA=L
SO,
-66=-(2L^2+16L)
SO,
2L^2+16L=66
2L^2+16L-66=0
2L^2+22L-6L-66=0
2L(L+11)-6(L+11)=0
(2L-6)(L+11)=0
2L-6=0 or L+11=0
2L=6 or L=(-11)
L=6/2=3 or L=(-11)
Let's take L=3
So, RA=3.......eqn3
(R+A)^2=(R-A)^2+4RA
=(2^2)+(4×3)
=4+12=16
So,
R+A=*16=4...eqn4
Add eqn2+eqn4
R-L+R+L=2+4
2R=6
R=6/2=3
L=4-R=4-3=1
Now L=1
So,
{(X+2)(2x+11)}/(x/4)=1
(X+2)(2x+11)=x+4
2x^2+15x+22=x+4
2x^2+15x-x+22-4=0
2x^2+14x+18=0
2(x^2+7x+9)=0
X^2+7x+9=0
Here a=1, b=7,c=9
D=b^2-4acc
=(7^2)-(4×1×9)
=49-36=13
*D=*13
X=(-b±*D)/2a
={(-7)±*13)/2
Hence x={(-7)±*13}/2
If we take L=(-11)
So, RA=(-11)
(R+A=)*{(2^2)-44}
=*(4-44)=*(-40)=2.*10i... Eqn5
Eqn2 +eqn5
R-A+R+A=2+(2.*10i)
2R=2(1+*10i)
R=1+*10i... Which provide a complex solution

[@RealQinnMalloryu4]

(x^4+24)(8x^4+20)/{x^4+16{+(x^4+8)(8x^4+44)/{x^4+16}={24x^4+28x^4}/16x^4 {8x^4+52x^4}/16x^4={52x^8/16x^4+60x^8/16x^4}=112x^16/32x^8=311x^2 3x^2(x ➖ 3x+2).2^42 2^21 2^3^7 2^3^3^4 2^3^3^2^2 1^1^3^1^2 32(x ➖ 3x+2).

[@HulkarxonQodirova]

It is very easy 😂😂