There is much simpler: X^4 = 15X - 14 X^4-X-14X-14=0 X(X^3-1)-14(X-1)=0 X(X-1)(X²+X+1)-14(X-1)=0 (X-1)[X(X²+X+1)-14]=0 (X-1)(X^3+X²+X-14)=0 2^3+2²+2-14=0 (X-1)(X-2)(X²+3X+7)=0 X=1 or X=2 are the only real solutions. there's also two complex solutions.
instead subtract 1 from both sides. Thats more simpler.x^4-1 = 15x-15. (x^2)^2-(1)^2 = 15(x-1).(x^2+1)(x^2-1) = 15(x-1). (x^2+1)(x+1)(x-1) = 15(x-1). taking everything on lhs (x^2+1)(x+1)(x-1)-15(x-1) = 0 (x-1){(x^2+1)(x+1)-15} = 0. x-1 = 0 so x = 1 is one solution. x^3+x^2+x-14 = 0. Put x=2. which makes 2^3+2^2+2-14 = 8+4+2-14 = 0 so x = 2 is other solution. For remaining two solution divide by x-2 and resolve.
Great solution, that's a clever approach to simplifying the equation! 👍💯😊Thanks for the awesome breakdown! It's always helpful to see different methods for solving equations! 🔥🔥🔥
Yes, great. After losing factor (x-1), instead of polynomial division by (x-2), one could share the -14 as powers of 2: x^3+x^2+x-14 = x^3-8 + x^2-4 +x-2 = 0 you see difference of cubes plus difference of squares plus direct difference of x and 2, hence: 0 = (x-2) ( (x^2+2x+4) + (x+2) + 1 ) = (x-2)( x^2 + 3x+ 7)
We can see x=1 isa root by synthetic division we get other roots are given by a cubic polynomial which gives 2 as a root . Again by synthetic division other roots are given by a.quadratic which gives complex roots. So the real roots are 1,2
There is much simpler:
X^4 = 15X - 14
X^4-X-14X-14=0
X(X^3-1)-14(X-1)=0
X(X-1)(X²+X+1)-14(X-1)=0
(X-1)[X(X²+X+1)-14]=0
(X-1)(X^3+X²+X-14)=0 2^3+2²+2-14=0
(X-1)(X-2)(X²+3X+7)=0
X=1 or X=2 are the only real solutions. there's also two complex solutions.
Spectacular.
One can see at once that 1 and 2 are solurtion. Then one can divide x^4-15x+14 by (x-1)(x-2).
x=1 and x=2 are roots by rational root theorem
hence x^2-3x+2 is a factor
by polynomial division other factor is x^2+3x+7
other method
Let
x^4- 15 x+14 = ( x^2+ax+b)(x^2- ax+ 14/b)
a (14/b-b) = -15
14/b +b -a^2 = 0
14/b+b = a^2
14/b-b = -15/a
a^4 - 225/a^2 = 56
Let a^2= t
t^3-56t -225 = 0
t = 9 hence a= 3
b+14/b= 9
14/b-b = -5
b= 7
(x^2+3x+7)(x^2-3x+2) = 0
instead subtract 1 from both sides. Thats more simpler.x^4-1 = 15x-15. (x^2)^2-(1)^2 = 15(x-1).(x^2+1)(x^2-1) = 15(x-1). (x^2+1)(x+1)(x-1) = 15(x-1). taking everything on lhs (x^2+1)(x+1)(x-1)-15(x-1) = 0
(x-1){(x^2+1)(x+1)-15} = 0. x-1 = 0 so x = 1 is one solution. x^3+x^2+x-14 = 0. Put x=2. which makes 2^3+2^2+2-14 = 8+4+2-14 = 0 so x = 2 is other solution. For remaining two solution divide by x-2 and resolve.
Great solution, that's a clever approach to simplifying the equation! 👍💯😊Thanks for the awesome breakdown! It's always helpful to see different methods for solving equations! 🔥🔥🔥
Yes, great. After losing factor (x-1), instead of polynomial division by (x-2), one could share the -14 as powers of 2:
x^3+x^2+x-14 = x^3-8 + x^2-4 +x-2 = 0
you see difference of cubes plus difference of squares plus direct difference of x and 2, hence:
0 = (x-2) ( (x^2+2x+4) + (x+2) + 1 ) = (x-2)( x^2 + 3x+ 7)
We can see x=1 isa root by synthetic division we get other roots are given by a cubic polynomial which gives 2 as a root . Again by synthetic division other roots are given by a.quadratic which gives complex roots. So the real roots are 1,2
Great solution! Thanks for the awesome breakdown! Thanks for the insightful explanation! 👍
In such problems if one (or two) roots are obvious, then it is easier to use conventional way. In this case 1 and 2 are obvious roots.
another method
Since there are no x^3 and x^2 terms, given equation can be expressed as followed:
x^4 - 15x + 14 = (x^2 + a/2)^2 - a(x + b/2)^2 = x^4 - abx + (a^2 - ab^2)/4
=> -ab = -15 (ab = 15 ), (a^2 - ab^2)/4 = 14 (a^2 - ab^2 = 56)
=> a^2 - ab^2 = a^2 - b(ab) = a^2 - 15b = 56 => b = (a^2 -56)/15
=> ab = 15 => a*(a^2 -56)/15 = 15 => a^3 - 56a - 225 = 0 => a = 9 => b = 5/3
=> x^4 - 15x + 14 = (x^2 + a/2)^2 - a(x + b/2)^2 = (x^2 + 9/2)^2 - 9(x + 5/6)^2
= (x^2 + 9/2 + 3x + 5/2)*(x^2 + 9/2 - 3x - 5/2) = (x^2 + 3x + 7)(x^2 - 3x + 2) = 0
Thanks for the alternative approach to the problem 🥰🙏💕💯🤩
The sum of the coefficients on the right-hand side is the same as the coefficient on the left-hand side. This implies that x=1 is a solution.
The equation has two roots 1 and 2, by division you get x^2+3x+7=0 you get the same roots, very easier.
X*(x^3-1)= 13*(x-1)
(x-1)*(x*(x^2+x+1)-13)=0
Dolution 1:
x = 1
Solution 2:
x^3 + x^2 +x -13=0
F(x)x^4 = 4x^3
F'(x)4x^3 = 12x^2
F"(x)12x^2 = 24x
24x - 15x = 9x
9x = - 14
x = - 14 / 9
x = - 1.55555
Demostrate.
- 1.55555*4*3*2 = 15(- 1.55555) - 14
- 37.33325 = - 37.33325
x⁴ = 15x - 14
x⁴ - 15x + 14 = 0 ← as the equation is quite simple in terms of coefficients, it's always a good idea to look for an obvious root (- 2 ; - 1 ; 1 ; 2)
x⁴ - 15x + 14 = 0 ← you can see thet 1 is an obvious root, so we can factirze (x - 1)
(x - 1).(x³ + ax² + bx - 14) = 0 → you expand
x⁴ + ax³ + bx² - 14x - x³ - ax² - bx + 14 = 0 → you group
x⁴ + x³.(a - 1) + x².(b - a) - x.(14 + b) + 14 = 0 → you compare with: (x⁴ - 15x + 14) = 0
For x³: → (a - 1) = 0 → a = 1
For x²: → (b - a) = 0 → b = a → b = 1
For x: → (14 + b) = 15 → b = 1 ← of course because above
(x - 1).(x³ + ax² + bx - 14) = 0 → where: a = b = 1
(x - 1).(x³ + x² + x - 14) = 0
First case: (x - 1) = 0
x - 1 = 0
→ x = 1 ← this is the obvious root
Second case: (x³ + x² + x - 14) = 0
x³ + x² + x - 14 = 0 ← the equation is quite simple once again, let's look for an obvious root (- 2 ; - 1 ; 1 ; 2)
x³ + x² + x - 14 = 0 ← you can see thet 2 is an obvious root, so we can factirze (x - 2)
(x - 2).(x² + ax + 7) = 0 → you expand
x³ + ax² + 7x - 2x² - 2ax - 14 = 0 → you group
x³ + x².(a - 2) + x.(7 - 2a) - 14 = 0 → you compare with: (x³ + x² + x - 14) = 0
For x²: → (a - 2) = 1 → a = 3
For x: → (7 - 2a) = 1 → 2a = 6 → a = 3 ← of course because above
(x - 2).(x² + ax + 7) = 0 → where: a = 3
(x - 2).(x² + 3x + 7) = 0
First case: (x - 2) = 0
x - 2 = 0
→ x = 2 ← this is the second obvious root
Second case: (x² + 3x + 7) = 0
x² + 3x + 7 = 0
Δ = 3² - (4 * 7) = 9 - 28 = - 19 = 19i²
→ x = (- 3 ± i√19)/2
Thanks for your powerful input 👍💯😊😍🤩🙏
(X^4 -X)-(14X-14)=0 X(X^3-1) -14(x-1)=0 (X-1)((X(X^2+X+1)-14)=0 (X-1)(X-2)((X^2 +2X +4 )+(X+2) +1)=0 (X-1)(X-2)(X^2-3X+7)=0
Apply logerthems, very simple, with simple multiplications.
Glad you found it simple. It's a fun challenge for sure! 🙏💕😎💯🥰
Logerthems ?
Jesus, where does the K & M come from ?
Watch and learn 😎💯💕😍🤩
x^4 +/- x^3 +/- x^2 - 15x + 14 = 0 , (x-2)(x^3+2x^2+4x-7)=0 , x^3+2x^2+4x-7=0 , (x-1)(x^2+3x+7)=0 ,
1 -2 1 -1 x^2+3x+7=0 , x=(-3+/-V(9-28))/2 ,
2 -4 3 -3 x=(-3+/-i*V(19))/2 ,
4 -8 7 -7
-7 14
(x^4)^2 ➖ (15)^2 ➖ 14 {x^16 ➖ 225}➖ 14 209 ➖ (14)^2={209 ➖ 196}=3 (x ➖ 3x+3).
Another method
x⁴=15x-14
x⁴-15x+14=0
x=1 is root?
| 1 | 0 | 0 |-15| 14|
1 | | 1 | 1 | 1|-14|
| 1 | 1 | 1 |-14| 0| YES
(x-1)(x³+x²+x-14)=0
x=2 is root?
| 1 | 1 | 1 |-14|
2 | | 2 | 6 | 14|
| 1 | 3 | 7 | 0| YES
(x-1)(x-2)(x²+3x+7)=0 ===> x-1=0 or x-2=0 or x²+3x+7=0
x=(-3±√(9-28))/2=(-3±√(-19))/2=(-3±i√19)/2
Finaly
x₁ = 1
x₂ = 2
x₃ = (-3+i√19)/2
x₄ = (-3-i√19)/2