I used synthetic division to break down the quartic, don't forget to have the zero place holder where the x^3 value should be and your golden. Once down to the quadratic, complete the square to find the last two possible solutions. As P and Q are small integer values , finding the factors (1,-2) in short order shouldn't be an issue with the factor theorem. You can do a quick check with Descartes rule of signs to find all solutions are real. Then plug in your values to check solutions. Apologies for getting the order all jumbled up, do Descartes rule of signs first, then the synthetic division.
The conditions you set at the beginning only apply if you are limiting the solutions to the real numbers. Imaginary solutions are perfectly valid generally. A quartic equation should have four roots and typically two are imaginary.
Except the original equation is not a quartic. It only becomes one because we square both sides to facilitate solving, thus creating the possibility of two extraneous solutions. Note also that neither of the solutions that were rejected [x=1 and x=(1-√13)/2] are imaginary, they really are just extraneous solutions that do not apply to the original equation. For example, if you plug x=1 into x²-3=√(x+3), you end up with this obviously faulty statement: -2=2. If instead you plug in (1-√13)/2, you end up with similar nonsense.
But... -why- did you decide t=3. Watching this, the algebra was set out slowly but the magic step was just "let's now do this inexplicably clever thing that happens to work" and never touched on again
Your solution is not a math reasoning: your choice if t implies that you already know the solution x=-2. You had no reason for this choice unless you knew the answer. If you knew the answer your bla bla bla is waisting our time.
C'est ma propre solution, tellement plus simple et convaincante ! Au plan des IDEES : 1. Notez que SEULE la condition x^2 - 3 > 0 est nécessaire et élimine les SOLUTIONS ETRANGERES. La condition "x + 3 > 0" est AUTOMATIQUEMENT satisfaite par l'élévation au carré. 2. Quant aux solutions entières apparentes du polynôme x4 - 6x2 - x - 6, elles doivent NECESSAIREMENT diviser (au signe près) le coefficient -6. D'où la factorisation par (x-1) puis (x+2).
x+3 ≥ 0 ---(1) x²-3 ≥ 0---(2) (1)---> x ≥ -3 => x = -3 -2, -1, -0, 1, 2, ... So on (2) ---> x² ≥ 3 => x² = 4, 5, 6, 7, 8, 9, 10, ....so on Let x = -3 (-3)²-3=9-3=6≠√(-3+3)=0 x≠-3 Let x = -2 (-2)²-3=4-3=1=√(-2+1)=√1=1 So, x = -2
way to complicated and I suppose, the substitution with t is not really needed. after using binomial theorem to expand (x^2-3)^2 and some substractions you get x^4-6x^2−x+6=0. This is a polynomial that is easy to solve. .
It does work! .. x⁴-x-6x²+6=x(x³-1)-6(x²-1)= = x(x-1)(x²+x+1)-6(x-1)(x+1)= =(x-1)[x(x²+x+1)-6(x+1)]= =(x-1)(x³+x²-5x-6)= =(x-1)(x³+2x²-x²-2x-3x-6)= =(x-1)[x²(x+2)-x(x+2)-3(x+2)]= =(x-1)(x+2)(x²-x-3)=0 => x=1 or x=-2 or x=(1±√13)/2. Due to constraints x=1 and x=(1-√13)/2 rejected .. Solutions x=-2 or x=(1+√13)/2.
I used synthetic division to break down the quartic, don't forget to have the zero place holder where the x^3 value should be and your golden. Once down to the quadratic, complete the square to find the last two possible solutions. As P and Q are small integer values , finding the factors (1,-2) in short order shouldn't be an issue with the factor theorem.
You can do a quick check with Descartes rule of signs to find all solutions are real. Then plug in your values to check solutions. Apologies for getting the order all jumbled up, do Descartes
rule of signs first, then the synthetic division.
You made a mistake in your final check, it should be sgrt(-2+3) not sqrt(-2+1), which would have created i, not sqrt(1) = 1.
(x^2 - 3)^2 = x+3
(x^2 - 3)^2 - x^2 = x+3 - x^2
(x^2 - 3+x) (x^2 - 3- x)+(x^2 - x- 3) = 0
(x^2 - x - 3) (x^2+x-2) = 0
(x^2 - x -3) (x+2)(x - 1) = 0
x = (1+√13)/2 , (1 - √13)/2 , - 2 , 1
x = (1+√13)/2 , - 2
other two roots are extraneous.
The conditions you set at the beginning only apply if you are limiting the solutions to the real numbers. Imaginary solutions are perfectly valid generally. A quartic equation should have four roots and typically two are imaginary.
Except the original equation is not a quartic. It only becomes one because we square both sides to facilitate solving, thus creating the possibility of two extraneous solutions. Note also that neither of the solutions that were rejected [x=1 and x=(1-√13)/2] are imaginary, they really are just extraneous solutions that do not apply to the original equation. For example, if you plug x=1 into x²-3=√(x+3), you end up with this obviously faulty statement: -2=2. If instead you plug in (1-√13)/2, you end up with similar nonsense.
But... -why- did you decide t=3. Watching this, the algebra was set out slowly but the magic step was just "let's now do this inexplicably clever thing that happens to work" and never touched on again
that piece of hair got me 😅
When you are afraid of a question by appearance. Simple factor would have made it so much more easier....
Your solution is not a math reasoning: your choice if t implies that you already know the solution x=-2. You had no reason for this choice unless you knew the answer. If you knew the answer your bla bla bla is waisting our time.
That's pretty the same I felt....
we get , x^4+/-x^3-6x^2-x+6=0 , (x-1)(x^3+x^2-5x-6)=0 , / x=1 , not a solu , /
1 -1 x^3+x^2-5x-6=0 , (x+2)(x^2-x-3)=0 , x= -2 , test , (-2)^2-3=1 , V(-2+3)=1 , same , OK ,
1 -1 1 2 x^2-x-3=0 , x= (1+V13)/2 , / (1-V13)/2 , not a solu / , ,
-5 5 -1 -2 test x=(1+V13)/2 , (1+V13)/2=(1+V13)/2 , same , OK ,
-6 6 -3 -6 solu , x= -2 , (1+V13)/2 ,
x^2 - 3 = √(x + 3)
x^4 - 6x^2 + 9 = x + 3
x^4 - 6x^2 - x + 6 = 0
(x - 1)(x^3 + x^2 - 5x - 6) = 0
(x - 1)(x + 2)(x^2 - x - 3) = 0
x = -2, (1 +/- √13)/2
C'est ma propre solution, tellement plus simple et convaincante !
Au plan des IDEES :
1. Notez que SEULE la condition x^2 - 3 > 0 est nécessaire et élimine les SOLUTIONS ETRANGERES. La condition "x + 3 > 0" est AUTOMATIQUEMENT satisfaite par l'élévation au carré.
2. Quant aux solutions entières apparentes du polynôme x4 - 6x2 - x - 6, elles doivent NECESSAIREMENT diviser (au signe près) le coefficient -6. D'où la factorisation par (x-1) puis (x+2).
Error when typing : X4 - 6x2 - x +6 , of course...!
@@thierrygermain5182 👌👍😊
Hi I like your method however I did not understand the substitution with t =3 can you please help
Thanks
Hello. Can you show how to solve this equation without using the Lambert function? 3^(x-2)=x
(x^2)^2 ➖ (3)^2={x^4 ➖ 9}=x^5;x^2^3 (x ➖ 3x+2).{ x+x ➖ }+{3+3 ➖ }={x^2+6}=6x^2 3^3x^2 1^1x^2 1x^2 (x ➖ 2x+1).
I just solved the quartic in x then tested the solns against the original rejected 2 roots kept 2
We already did it with one and four and five
Thank
I got this question in my olympiad exam and i didn't know how to solve it : [2^(x-3)] × [3^(2x-8) ]=36 ; find x
1 is an obvious solution?
1 - 3 = -2
x+3 ≥ 0 ---(1)
x²-3 ≥ 0---(2)
(1)---> x ≥ -3 => x = -3 -2, -1, -0, 1, 2, ... So on
(2) ---> x² ≥ 3 => x² = 4, 5, 6, 7, 8, 9, 10, ....so on
Let x = -3
(-3)²-3=9-3=6≠√(-3+3)=0
x≠-3
Let x = -2
(-2)²-3=4-3=1=√(-2+1)=√1=1
So, x = -2
I got - 2
We are runnung out of figures
Are you spanish?
way to complicated and I suppose, the substitution with t is not really needed. after using binomial theorem to expand (x^2-3)^2 and some substractions you get x^4-6x^2−x+6=0. This is a polynomial that is easy to solve.
.
Trop long et compliqué.
Est ce qu'il n'y avait pas moyen de faire simple?
Merci pour les habilités.
√(x + 3) = y
x + 3 = y²
x² - 3 = y
x² + x = y² + y
(x² - y²) + (x - y) = 0
(x - y)(x + y + 1) = 0
y = x ∨ y = -(x + 1)
y = x
x² - 3 = x => x² - x - 3 = 0
x = (1 ± √13)/2
*x = (1 + √13)/2* ∨ x = (1 - √13)/2 (rejected)
y = -(x + 1)
x² - 3 = -(x + 1)
x² + x - 2 = 0
x = (-1 ± 3)/2
x = 1 (rejected) ∨ *x = -2*
ANOTHER WAY
x² = √(x + 3) + 3
x⁴ = x + 12 + 6√(x + 3)
x⁴ = x + 12 + 6(x² - 3)
x⁴ - 6x² - x + 6 = 0
x(x³ - 1) - 6(x² - 1) = 0
x(x - 1)(x² + x + 1) - 6(x - 1)(x + 1) = 0
x - 1 ≠ 0 => x³ + x² + x - 6x - 6 = 0
x³ + x² - 5x - 6 = 0
*x = -2* is a solution (by inspection)
(x + 2)(x² - x - 3) = 0
x² - x - 3 = 0 => x = (1 ±√13)/2
*x = (1 + √13)/2* ∨ x = (1 - √13)/2 (rejected)
It does work!
.. x⁴-x-6x²+6=x(x³-1)-6(x²-1)=
= x(x-1)(x²+x+1)-6(x-1)(x+1)=
=(x-1)[x(x²+x+1)-6(x+1)]=
=(x-1)(x³+x²-5x-6)=
=(x-1)(x³+2x²-x²-2x-3x-6)=
=(x-1)[x²(x+2)-x(x+2)-3(x+2)]=
=(x-1)(x+2)(x²-x-3)=0 =>
x=1 or x=-2 or x=(1±√13)/2.
Due to constraints x=1 and x=(1-√13)/2 rejected ..
Solutions x=-2 or x=(1+√13)/2.
@@gregevgeni1864 yes. I have fixed my "ANOTHER WAY". thanks.