The given equation simplifies to (x^2-x)/[sqrt(x^2-x+24)] =-2. Let t = x^2-x > t/[sqrt(t+24)] = -2 > t^2-4t-96=0 > t = -8,12.If t = -8, x^2-x+8=0, which has no real roots. If t=12, x^2-x-12 =0 > x=-3,4. We can readily verify that x=-3 is a spurious solution and that x=4 is the only real solution.
After some manipulations, we have to solve the equation x⁴ - 2x³ - 3x² + 4x - 96=0, witch is (x-4)(x+3)(x² - x + 8)=0. Thus, x=4 and x=-3 as real solutions but x=-3 is rejected after checking in the original equation, and two complex ones x=(1+or-isqrt31)/2.
Multiplying LHS by (x / x) and simplifying
√(x² - 2x + 1)/√(x² - x + 24) = 2 / x
=> x√(x - 1)² = 2√(x² - x + 24)
=> x² - x = 2√(x² - x + 24)
Substituting t = √(x² - x + 24)
=> t² - 24 = 2t
=> t² - 2t - 24 = 0
t = 6, -4 and x² - x + 24 = t²
So, x² - x + 24 = 36 or x² - x + 24 = 16
x² - x -12 = 0 ( x² - x + 8 = 0 reject)
x = 4 (reject x = -3)
The given equation simplifies to (x^2-x)/[sqrt(x^2-x+24)] =-2. Let t = x^2-x > t/[sqrt(t+24)] = -2 > t^2-4t-96=0 > t = -8,12.If t = -8, x^2-x+8=0, which has no real roots. If t=12, x^2-x-12 =0 > x=-3,4. We can readily verify that x=-3 is a spurious solution and that x=4 is the only real solution.
After some manipulations, we have to solve the equation x⁴ - 2x³ - 3x² + 4x - 96=0, witch is (x-4)(x+3)(x² - x + 8)=0. Thus, x=4 and x=-3 as real solutions but x=-3 is rejected after checking in the original equation, and two complex ones x=(1+or-isqrt31)/2.
X= 4; -3 x #
Στο συνολο R χ=-3 ή χ=4. Στο συνολο των μιγαδικων επι πλεον οι ριζες χ=(1+ -ριζα31)/2
X=-3,4,((1+4√2 i)/2),(1-4√2 i/2))
1/x=t
(1)^2 ➖ (2)^2/(x)^2= {1 ➖ 4}/x^2=3/x^2 {1x+1x}/{x+x ➖} 2x^2/x^2 {3/x^2+2x^2/x^2}=6x^2/x^4 1.2x^2 1.1x2 1x^2 (x ➖ 2x+1) (1)^2=1 (1)^2/(x)^2 = 1/x^2 {1 ➖ 1/x^2}= 0+0 ➖/x^2= 1/x^2 {24x+24x ➖}/{x+ x ➖} = 48x^2/x^2 {1/x^2+48x^2/x^2}=49x^2/x^4 =12.x^2 3^4 x^2 3^2^2 x^2 3^1^1x^2 3x^2 (x ➖ 3x+2)
Multiply with x ( x>0) you get
Sqrt((x-1)^2) = 2*/x * sqrt( x^2-x+24)
(X-1)^2 *x^2 = 4* ( x^2 - x +24)
Therefore x^4 -2x^3-3x^2+4x -96=0. You find x=4 , conclude
( x-4)* ( x^3+ 2x^2+5x +24) =0
Second solution x= -3