Rewrite as x^3 -x^2 + 2x - 24 = 0. By rational root theorem, any possible rational roots would come from (+/-) 1, 2, 3, 4, 6, 8, 12, or 24. A little trial and error lead to +3 as a rational solution. By synthetic division, you can rewrite x^3 -x^2 + 2x - 24 as (x -3) (x^2 + 2x +8). The roots of x^2 + 2x +8 can be found with the quadratic formula. So the three solutions are x = 3 or x = 1 ± i√7)
x^3-x^2+2x-24=0
(x-3)(x^2+2x+8)=0
x=3 or x=-1+irt7
or x=-1-irt7
Rewrite as x^3 -x^2 + 2x - 24 = 0. By rational root theorem, any possible rational roots would come from (+/-) 1, 2, 3, 4, 6, 8, 12, or 24. A little trial and error lead to +3 as a rational solution. By synthetic division, you can rewrite x^3 -x^2 + 2x - 24 as (x -3) (x^2 + 2x +8). The roots of x^2 + 2x +8 can be found with the quadratic formula. So the three solutions are x = 3 or x = 1 ± i√7)
Yes, I can.
not 14.50 but half of minutes
x³ - x² + 2x = 24
x³ - x² + 2x - 24 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the power 2
Let: x = z - (b/3a) → where:
b is the coefficient for x², in our case: - 1
a is the coefficient for x³, in our case: 1
x³ - x² + 2x - 24 = 0 → let: x = z - (- 1/3) → x = z + (1/3)
[z + (1/3)]³ - [z + (1/3)]² + 2.[z + (1/3)] - 24 = 0
[z + (1/3)]².[z + (1/3)] - [z² + (2/3).z + (1/9)] + 2z + (2/3) - 24 = 0
[z² + (2/3).z + (1/9)].[z + (1/3)] - z² - (2/3).z - (1/9) + 2z + (2/3) - 24 = 0
[z³ + (1/3).z² + (2/3).z² + (2/9).z + (1/9).z + (1/27)] - z² - (2/3).z - (1/9) + 2z + (2/3) - 24 = 0
z³ + z² + (1/3).z + (1/27) - z² - (2/3).z - (1/9) + 2z + (2/3) - 24 = 0
z³ + (5/3).z - (632/27) = 0 ← no term to the power 2
z³ + (5/3).z - (632/27) = 0 → let: z = u + v
(u + v)³ + (5/3).(u + v) - (632/27) = 0
(u + v)².(u + v) + (5/3).(u + v) - (632/27) = 0
(u² + 2uv + v²).(u + v) + (5/3).(u + v) - (632/27) = 0
(u³ + u²v + 2u²v + 2uv² + uv² + v³) + (5/3).(u + v) - (632/27) = 0
u³ + v³ + 3u²v + 3uv² + (5/3).(u + v) - (632/27) = 0
u³ + v³ + 3uv.(u + v) + (5/3).(u + v) - (632/27) = 0
u³ + v³ + (u + v).[3uv + (5/3)] - (632/27) = 0 → supose that: [3uv + (5/3)] = 0 ← equation (1)
u³ + v³ + (u + v).[0] - (632/27) = 0
u³ + v³ - (632/27) = 0 ← equation (2)
Then you get a system of 2 equations
(1): [3uv + (5/3)] = 0
(1): 3uv = - 5/3
(1): uv = - 5/9
(1): u³v³ = - (5/9)³ ← this is the product P
(2): u³ + v³ - (632/27) = 0
(2): u³ + v³ = 632/27 ← this is the sum S
u³ & v³ are the solution of the equation:
a² - Sa + P = 0
a² - (632/27).a - (5/9)³ = 0
Δ = (- 632/27)² - 4.[1 * - (5/9)³] = (632²/27²) + (500/729) = 399924/27² = (138²/27²) * 21
a = [(632/27) ± (138/27)√21]/2
a = (316/27) ± (69/27)√21
a = (316 ± 69√21)/27
u³ = (316 + 69√21)/27 → u = [(316 + 69√21)/27]^(1/3)
v³ = (316 - 69√21)/27 → v = [(316 - 69√21)/27]^(1/3)
Recall: z = u + v
Recall: x = z + (1/3)
x = [(316 + 69√21)/27]^(1/3) + [(316 - 69√21)/27]^(1/3) + (1/3)
x = 3
Resart from the equation:
x³ - x² + 2x - 24 = 0 → we've seen that 3 is a root, so you can factorize: (x - 3)
(x - 3).(x² + ax + 8) = 0 → you expand
x³ + ax² + 8x - 3x² - 3ax - 24 = 0 → you group
x³ + x².(a - 3) + x.(8 - 3a) - 24 = 0 → you compare with: x³ - x² + 2x - 24 = 0
For x² → (a - 3) = - 1 → a = 2
For x → (8 - 3a) = 2 → - 3a = - 6 → a = 2 (of course because above)
(x - 3).(x² + ax + 8) = 0 → where: a = 2
(x - 3).(x² + 2x + 8) = 0
First case: (x - 3) = 0 → x = 3 ← this is the first root calculated according to Cardan' method
Second case: (x² + 2x + 8) = 0
x² + 2x + 8 = 0
Δ = (2)² - (4 * 8) = 4 - 32 = - 28 = 28i²
x = (- 2 ± i√28)/2
x = (- 2 ± 2i√7)/2
x = - 1 ± i√7
x = - 1 + i√7 ← this is the second root
x = - 1 - i√7 ← this is the third root
Thanks for detailed explanation 🙏💕💯🥰✅💪
Cambridge University Entrance Exam: x³ - x² + 2x = 24; x =?
x³ - x² + 2x - 24 = 0, x³ - 27 - x² + 2x + 3 = (x³ - 3³) - (x² - 2x - 3) = 0
(x - 3)(x² + 3x + 9) - (x - 3)(x + 1) = (x - 3)(x² + 2x + 8) = 0
x - 3 = 0; x = 3 or x² + 2x + 8 = 0, (x + 1)² = - 7 = (i√7)²; x = - 1 ± i√7
Answer check:
x = 3: x³ - x² + 2x = x(x² - x + 2) = 3(9 - 3 + 2) = 3(8) = 24; Confirmed
x = - 1 ± i√7: x² + 2x + 8 = 0; x² = - 2x - 8
x³ - x² + 2x = x(x² - x + 2) = x(- 2x - 8 - x + 2) = - 3x(x + 2)
= - 3(- 1 ± i√7)(- 1 ± i√7 + 2) = - 3(- 1 ± i√7)(1 ± i√7) = - 3(- 8) = 24; Confirmed
Final answer:
x = 3; Two complex value roots, x = - 1 + i√7 or x = - 1 - i√7
x³-x²+2x=24; solution; x(x²-x+2)=24; x=24; x²-x+2=24;x²-x=24-2;x²-x=20; x²-x=20; x²-x-20;x²-x-16-4=0;x²-x-4²-4=0; x²-4²-x-4=0; (x-4)(x+4)-(x-4)=0; (x-4)[x+4-1]=0; x-4=0; x=4; x+3=0; x=-3; S={-3; 4;24}.
x^3-x^2+2x-24=0=(x+a)(x^2+bx+c);=> {ac=-24; abx+cx=2x;(ab+c)=2; ax^2+b^x^2=- x^2 (a+b=-1) => a=-3; b=2; c= 8;} =>Eqn.=> (x-3)(x^2+2x+8)=0