(x^2-9/x^2-3x-4)/(x^2+2x-15/x^2-x-12)
1 answer:
Eliminate the cubic term by substituting y = x - 3/4:-9 - 4 (y + 3/4)^2 - 3 (y + 3/4)^3 + (y + 3/4)^4 = 0
Expand out terms of the left hand side:y^4 - (59 y^2)/8 - (75 y)/8 - 3123/256 = 0
Subtract -(75 y)/8 - (59 y^2)/8 - 1/8 (3 i) sqrt(347) y^2 from both sides:-3123/256 + 1/8 (3 i) sqrt(347) y^2 + y^4 = (75 y)/8 + (59 y^2)/8 + 1/8 (3 i) sqrt(347) y^2
-3123/256 + 1/8 (3 i) sqrt(347) y^2 + y^4 = (y^2 + 1/16 (3 i) sqrt(347))^2:(y^2 + 1/16 (3 i) sqrt(347))^2 = (75 y)/8 + (59 y^2)/8 + 1/8 (3 i) sqrt(347) y^2
Add 2 (y^2 + 1/16 (3 i) sqrt(347)) λ + λ^2 to both sides:(y^2 + 1/16 (3 i) sqrt(347))^2 + 2 λ (y^2 + 1/16 (3 i) sqrt(347)) + λ^2 = (75 y)/8 + 3/8 i sqrt(347) y^2 + (59 y^2)/8 + 2 λ (y^2 + 1/16 (3 i) sqrt(347)) + λ^2
(y^2 + 1/16 (3 i) sqrt(347))^2 + 2 λ (y^2 + 1/16 (3 i) sqrt(347)) + λ^2 = (y^2 + (3 i sqrt(347))/16 + λ)^2:(y^2 + (3 i sqrt(347))/16 + λ)^2 = (75 y)/8 + 3/8 i sqrt(347) y^2 + (59 y^2)/8 + 2 λ (y^2 + 1/16 (3 i) sqrt(347)) + λ^2
(75 y)/8 + 3/8 i sqrt(347) y^2 + (59 y^2)/8 + 2 λ (y^2 + 1/16 (3 i) sqrt(347)) + λ^2 = (59/8 + (3 i)/8 sqrt(347) + 2 λ) y^2 + (75 y)/8 + 3/8 i sqrt(347) λ + λ^2:(y^2 + (3 i sqrt(347))/16 + λ)^2 = y^2 (59/8 + (3 i)/8 sqrt(347) + 2 λ) + (75 y)/8 + 3/8 i sqrt(347) λ + λ^2
Complete the square on the right hand side:(y^2 + (3 i sqrt(347))/16 + λ)^2 = (y sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ) + 75/(16 sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ)))^2 + (4 (59/8 + 1/8 (3 i) sqrt(347) + 2 λ) (1/8 (3 i) sqrt(347) λ + λ^2) - 5625/64)/(4 (59/8 + 1/8 (3 i) sqrt(347) + 2 λ))
To express the right hand side as a square, find a value of λ such that the last term is 0.This means 4 (59/8 + 1/8 (3 i) sqrt(347) + 2 λ) (1/8 (3 i) sqrt(347) λ + λ^2) - 5625/64 = 1/64 (-5625 - 12492 λ + (708 i) sqrt(347) λ + 1888 λ^2 + (288 i) sqrt(347) λ^2 + 512 λ^3) = 0.Thus the root λ = -1/48 i (-59 i + 9 sqrt(347) + (736 i) (2/(4907 + 9 sqrt(335721)))^(1/3) - (4 i) 2^(2/3) (4907 + 9 sqrt(335721))^(1/3)) allows the right hand side to be expressed as a square.(This value will be substituted later):(y^2 + (3 i sqrt(347))/16 + λ)^2 = (y sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ) + 75/(16 sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ)))^2
Take the square root of both sides:y^2 + (3 i sqrt(347))/16 + λ = y sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ) + 75/(16 sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ)) or y^2 + (3 i sqrt(347))/16 + λ = -y sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ) - 75/(16 sqrt(59/8 + 1/8 (3 i) sqrt(347) + 2 λ))
Solve using the quadratic formula:y = 1/8 (sqrt(2) sqrt(59 + (3 i) sqrt(347) + 16 λ) + 4 sqrt(59/8 - 1/8 (3 i) sqrt(347) - 2 λ + 75/sqrt(118 + (6 i) sqrt(347) + 32 λ))) or y = 1/8 (sqrt(2) sqrt(59 + (3 i) sqrt(347) + 16 λ) - 4 sqrt(59/8 - 1/8 (3 i) sqrt(347) - 2 λ + 75/sqrt(118 + (6 i) sqrt(347) + 32 λ))) or y = 1/8 (4 sqrt(59/8 - 1/8 (3 i) sqrt(347) - 2 λ - 75/sqrt(118 + (6 i) sqrt(347) + 32 λ)) - sqrt(2) sqrt(59 + (3 i) sqrt(347) + 16 λ)) or y = 1/8 (-(sqrt(2) sqrt(59 + (3 i) sqrt(347) + 16 λ)) - 4 sqrt(59/8 - 1/8 (3 i) sqrt(347) - 2 λ - 75/sqrt(118 + (6 i) sqrt(347) + 32 λ))) where λ = 1/48 (-i) (-59 i + 9 sqrt(347) + (736 i) (2/(4907 + 9 sqrt(335721)))^(1/3) - (4 i) 2^(2/3) (4907 + 9 sqrt(335721))^(1/3))
Substitute λ = -1/48 i (-59 i + 9 sqrt(347) + (736 i) (2/(4907 + 9 sqrt(335721)))^(1/3) - (4 i) 2^(2/3) (4907 + 9 sqrt(335721))^(1/3)) and approximate:y = -2.36475 or y = -0.494947 - 1.13703 i or y = -0.494947 + 1.13703 i or y = 3.35465
Substitute back for y = x - 3/4:x - 3/4 = -2.36475 or y = -0.494947 - 1.13703 i or y = -0.494947 + 1.13703 i or y = 3.35465
Add 3/4 to both sides:x = -1.61475 or y = -0.494947 - 1.13703 i or y = -0.494947 + 1.13703 i or y = 3.35465
Substitute back for y = x - 3/4:x = -1.61475 or x - 3/4 = -0.494947 - 1.13703 i or y = -0.494947 + 1.13703 i or y = 3.35465
Add 3/4 to both sides:x = -1.61475 or x = 0.255053 - 1.13703 i or y = -0.494947 + 1.13703 i or y = 3.35465
Substitute back for y = x - 3/4:x = -1.61475 or x = 0.255053 - 1.13703 i or x - 3/4 = -0.494947 + 1.13703 i or y = 3.35465
Add 3/4 to both sides:x = -1.61475 or x = 0.255053 - 1.13703 i or x = 0.255053 + 1.13703 i or y = 3.35465
Substitute back for y = x - 3/4:x = -1.61475 or x = 0.255053 - 1.13703 i or x = 0.255053 + 1.13703 i or x - 3/4 = 3.35465
Add 3/4 to both sides:Answer: x = -1.61475 or x = 0.255053 - 1.13703 i or x = 0.255053 + 1.13703 i or x = 4.10465
You might be interested in
Answer:
64 pi
Step-by-step explanation:
lmk if you need an explanation
Answer:
18x+28y
Step-by-step explanation:
The answer is A one triangle all those angles equal 180
Answer: the answer is A
Step-by-step explanation:
Answer: There are no solution
Step 1: Simplify both sides of the equation.
3(x+5)−3=2(3x+1)−3x
(3)(x)+(3)(5)+−3=(2)(3x)+(2)(1)+−3x(Distribute)
3x+15+−3=6x+2+−3x
(3x)+(15+−3)=(6x+−3x)+(2)(Combine Like Terms)
3x+12=3x+2
3x+12=3x+2
Step 2: Subtract 3x from both sides.
3x+12−3x=3x+2−3x
12=2
Step 3: Subtract 12 from both sides.
12−12=2−12
0=−10
