My solution Given equation:- √(6-√(6+x))=x √(6-√(6+x))=√(x)² Now comparing the radical inner term So,it becomes 6-√(6+x)=x² Subtracting '6' on both sides -√(6+x)=x²-6 Squaring on both sides we get 6+x=x⁴-12x²+36 Simply further x⁴-12x²-x+30=0 Now, P₁(x)=x⁴-12x²-x+30....(1) P₁(1)≠0 P₁(2)≠0 P₁(3)=0 So, x=3 is a root x-3 is a factor....(2) (1)÷(2) (x⁴-12x²-x+30)÷(x-3)=(x³+3x²-3x-10) So, P₂(x)=x³+3x²-3x-10....(3) Initial guess x=-2 P₂(-2)=0 So, x=-2 is a root x+2 is a factor....(4) (3)÷(4) (x³+3x²-3x-10.)÷(x+2)=(x²+x-5) P₃(x)=x²+x-5 We can use quadratic formula x=(-1±√21)÷(2) After some checking valid solution of 'x' x=(-1+√21)÷(2)
Good method. But I solved it using Syn, Div. once I formed my 4th degree polynomial, I found the factoring too long and the concept should match the level as factoring is a low level and your example is an upper level of algebra.
√[6 - √(6 + x)] = x x ≥ 0 √(6 + x) = u √(6 - u) = x 6 + x = u² 6 - u = x² u² - x² = u + x (u + x)(u - x) - (u + x) = 0 (u + x)(u - x - 1) = 0 u = - x ∨ u = x + 1 u = -x => 6 + x = x² x² - x - 6 = 0 (x - 3)(x + 2) = 0 x = 3 ∨ x = -2 [ both are invalid ] u = x + 1 => 6 - (x + 1) = x² x² + x - 5 = 0 *x = (-1 + √21)/2*
To solve the equation x^4-12x^2-x+30=0, first search for integer solutions. These must be integer factors (positive or negative) of 30. It is clear that the author knows that two solutions are x=3 and x=-2, and organizes the calculation to factor out (x-3)(x+2). This can be done by long polynomial division, which is essentially what the author does.
Почему 3 не подходит для решения? 6+3=9 корень из 9 = +3, - 3, берём-3 отнимаем от 6=9, корень из 9 = 3, -3 В правой части 3 И получается равно в 50%😂 Квантовая запутанность.
@SALogics Откуда вы взяли это ограничение? Проверка осуществляется постановкой на месте Х ответа и при постановке 3 вместо Х, нигде не получается корня из отрицательного числа.
@@ДмитрийЗенков-х7о If x² = 9, then x = 3 or x = - 3. But for ✓9 = 3 (always) not -3. So if x = 3, then ✓(x + 3) = ✓9 = 3, not -3. Please you make the graph of y = x and the curve of y = ✓(6 - ✓(x + 6)), it will intersect at only one point
My solution
Given equation:-
√(6-√(6+x))=x
√(6-√(6+x))=√(x)²
Now comparing the radical inner term
So,it becomes
6-√(6+x)=x²
Subtracting '6' on both sides
-√(6+x)=x²-6
Squaring on both sides we get
6+x=x⁴-12x²+36
Simply further
x⁴-12x²-x+30=0
Now,
P₁(x)=x⁴-12x²-x+30....(1)
P₁(1)≠0
P₁(2)≠0
P₁(3)=0
So,
x=3 is a root
x-3 is a factor....(2)
(1)÷(2)
(x⁴-12x²-x+30)÷(x-3)=(x³+3x²-3x-10)
So,
P₂(x)=x³+3x²-3x-10....(3)
Initial guess
x=-2
P₂(-2)=0
So,
x=-2 is a root
x+2 is a factor....(4)
(3)÷(4)
(x³+3x²-3x-10.)÷(x+2)=(x²+x-5)
P₃(x)=x²+x-5
We can use quadratic formula
x=(-1±√21)÷(2)
After some checking valid solution of 'x'
x=(-1+√21)÷(2)
Very nice trick! I really appreciate that ❤
This is great.
I like it.
Thanks for liking! ❤
To check that ((sqrt(21) -1)/2)^2
Very nice! ❤
Good method. But I solved it using Syn, Div. once I formed my 4th degree polynomial, I found the factoring too long and the concept should match the level as factoring is a low level and your example is an upper level of algebra.
Very nice! ❤
√[6 - √(6 + x)] = x
x ≥ 0
√(6 + x) = u
√(6 - u) = x
6 + x = u²
6 - u = x²
u² - x² = u + x
(u + x)(u - x) - (u + x) = 0
(u + x)(u - x - 1) = 0
u = - x ∨ u = x + 1
u = -x => 6 + x = x²
x² - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 ∨ x = -2 [ both are invalid ]
u = x + 1 => 6 - (x + 1) = x²
x² + x - 5 = 0
*x = (-1 + √21)/2*
Very nice trick! ❤
Very good!
You can immediately rule out x+u=0, since x>0, u>0. Otherwise very elegant solution❤.
Simple and a nice solution 👍🙏
Why can't we use factor theorum to solve the problem in simplest way.
Yes, it can be done so. ❤
x^4-12x^2-x+30=0
x=-2 16-48+2+30=0
this problem's solution is a bunch of hocus pocus. A lot of fumbling around to factor without any real structure.
To solve the equation x^4-12x^2-x+30=0, first search for integer solutions. These must be integer factors (positive or negative) of 30. It is clear that the author knows that two solutions are x=3 and x=-2, and organizes the calculation to factor out (x-3)(x+2). This can be done by long polynomial division, which is essentially what the author does.
a common task for a Russian school. a class with an in-depth study of mathematics. nothing complicated)
Very nice! ❤
Почему 3 не подходит для решения? 6+3=9 корень из 9 = +3, - 3, берём-3 отнимаем от 6=9, корень из 9 = 3, -3 В правой части 3 И получается равно в 50%😂 Квантовая запутанность.
3 отклоняется, поскольку квадрат 3 больше 6
и при проверке мы не можем брать -ve значение квадратного корня ❤
@SALogics Откуда вы взяли это ограничение? Проверка осуществляется постановкой на месте Х ответа и при постановке 3 вместо Х, нигде не получается корня из отрицательного числа.
@SALogics По существу это уравнения пересечение прямой и пораболы и прямая пересекает пораболу в 2 х точках.Нарисуйте график и всё станет понятно
@@ДмитрийЗенков-х7о
If x² = 9, then x = 3 or x = - 3.
But for ✓9 = 3 (always) not -3.
So if x = 3, then ✓(x + 3) = ✓9 = 3, not -3.
Please you make the graph of y = x and the curve of y = ✓(6 - ✓(x + 6)), it will intersect at only one point
sqrt[6-sqrt(6+x)]=x
square {sqrt[6-sqrt(6+x)]=x}
6-sqrt(6+x)=x^2
rearrange
x^2 - 6 = -sqrt(6+x)
squar{x^2 - 6 = -sqrt(6+x)}
6+x=[x^2-6]^2
/////
this results in an x^4 term yielding there's FOUR solutions.
6+x=x^4-12x^2+36
rearrange
x^4 +0x^3-12x^2 -x+30
Very nice trick! ❤
Plug 2 you get sqart of 6 - sqart sqart of 6+2 = sqart 6 - sqart sqart 8 = sqart 6-2=2
And x=2
Very nice! ❤
Схема Горнера
Very nice! ❤
Why not use synthetic division?
Nice Suggession! ❤
(6+X)^1/2=Y 6+X=Y^2 (1)
(6-Y)^1/2 =X 6-Y =X^2 (2) X>0 , X=0
(2) - (1) : X^2-Y^2+X+Y=0
(X+Y)(X-Y)+(X+Y) =0
(X+Y)(X-Y+1)=0
1. X=-Y (2): 6+X=X^2
X^2-X-6=0 X1=-2 , X2=3
X1=-2
Very nice trick! ❤
，
x = - 2 must rejected, because the condition is
0
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