My approach is by defining an intermediate variable based on the halfway point between the factors. For example, in (x+0)(x+2)(x+4)(x+6), 3 is halfway between 0 and 6, so I set y = x + 3. This gives a new equation (y-3)(y-1)(y+1)(y+3) = (y^2-1)(y^2-9) = (z+4)(z-4) = 9 where we set z = y^2 - 5. From this we solve backwards. z^2 - 16 = 9, so z = +/- 5. y^2 = z + 5, so y^2 = 10 or 0. x = y - 3, so x = -3, -sqrt(10)-3, +sqrt(10) -3.
03:31 I made it short and easy by using reduced quadratic equation formula a^2+px+q a_1=-(p/2)+sqrt((p/2)^2-q) a_2=-(p/2)-sqrt((p/2)^2-q) plug the values in: a_1=-4+sqrt(4^2+9)=1 a_2= -4-sqrt(4^2+9)=-9 07:47 and following: did the same there. Works fine. 👍😃 👍
Here is my solution which is easier to calculate: firstly: x(x+2)(x+4)(x+6)=x(x+6)(x+2)(x+4)=(x+3-3)(x+3+3)(x+3-1)(x+3+1)=[(x+3)^2-3^2][(x+3)^2-1^2] =9 let y=(x+3)^2, then we got: (y-9)(y-1)=y^2-10y+9=9, means: y(y-10)=0 so, y1=0, y2=10 now bring x back: in case: y1=(x+3)^2 = 0 => x=-3, in case: y2=(x+3)^2=10 => x+3 =±sqrt(10) => x=-3±sqrt(10)
You get a^2+8a'-9=0. Therefore (a -1)*(a+9) =0 then you have to Solve 2 easy quadratic equations a^2 +6x +9=0 conclude ( a +3)^2=0 a=-3 or x^2 +6x -1=0 easy with quadratic Formular. X= -3 + - sqrt( 10) thsts all
There are 4 terms that multiply to get 9. The only way to get 9 is if the terms are something like 3 x 3 x 1 x 1. To get (x + 6) to be one of the 3s, x = -3. The rest of the terms fit x(x+2)(x+4)(x+6) = -3(-1)(1)(3) = 9.
An excellent problem
My approach is by defining an intermediate variable based on the halfway point between the factors. For example, in (x+0)(x+2)(x+4)(x+6), 3 is halfway between 0 and 6, so I set y = x + 3. This gives a new equation (y-3)(y-1)(y+1)(y+3) = (y^2-1)(y^2-9) = (z+4)(z-4) = 9 where we set z = y^2 - 5.
From this we solve backwards. z^2 - 16 = 9, so z = +/- 5. y^2 = z + 5, so y^2 = 10 or 0. x = y - 3, so x = -3, -sqrt(10)-3, +sqrt(10) -3.
Awesome, I like that. This is fantastic.
Very clever 👏👏👏
@@davidbrisbane7206 Slight simplification to use that
(y^2-9)(y^2-1)-9=(y^2-10)y^2 = 0.
Otherwise elegant solution❤!
03:31 I made it short and easy by using reduced quadratic equation formula
a^2+px+q
a_1=-(p/2)+sqrt((p/2)^2-q)
a_2=-(p/2)-sqrt((p/2)^2-q)
plug the values in:
a_1=-4+sqrt(4^2+9)=1
a_2= -4-sqrt(4^2+9)=-9
07:47 and following: did the same there. Works fine. 👍😃 👍
That's fantastic.
Here is my solution which is easier to calculate:
firstly: x(x+2)(x+4)(x+6)=x(x+6)(x+2)(x+4)=(x+3-3)(x+3+3)(x+3-1)(x+3+1)=[(x+3)^2-3^2][(x+3)^2-1^2] =9
let y=(x+3)^2, then we got: (y-9)(y-1)=y^2-10y+9=9, means: y(y-10)=0
so, y1=0, y2=10
now bring x back:
in case: y1=(x+3)^2 = 0 => x=-3,
in case: y2=(x+3)^2=10 => x+3 =±sqrt(10) => x=-3±sqrt(10)
Set a=x+3 is a much easier substitution to work with.
Now i understand 😅
You get a^2+8a'-9=0. Therefore (a -1)*(a+9) =0 then you have to Solve 2 easy quadratic equations
a^2 +6x +9=0 conclude ( a +3)^2=0 a=-3 or x^2 +6x -1=0 easy with quadratic Formular.
X= -3 + - sqrt( 10) thsts all
все можно проще: z=x+3 (z-3)*(z-1)*(z+1)*(z+3)=9 (z^2-9)*(z^2-1)=9 z^2=t
(t--9)(t-1)=9 t^2-10t +9=9 t^2-10t=0
There are 4 terms that multiply to get 9. The only way to get 9 is if the terms are something like 3 x 3 x 1 x 1. To get (x + 6) to be one of the 3s, x = -3. The rest of the terms fit x(x+2)(x+4)(x+6) = -3(-1)(1)(3) = 9.
The only way? False, since x is not restricted to integers, or even the reals!
X=-3-√10 orX=-3+√10
X=-3
x=-3 is a repeated root ... a 4-th order equation must have 4 solutions.
Cool my man
x = -3
Too me 10 sec.
Idk why when I looked at the question I immediately thought of -3, tried it, and it worked lol
Bruh you had two cases where completing the square was easier then factorizing or the quadratic formula
t=x-3
(t-3)(t-1)(t+1)(t+3)=9
(t^2-9)(t^2-1)=9
t^4-10t^2+9=9
t^4-10t^2=0
t^2(t^2-10)=0
∴t=0,t=±√10
∴x=-3,-3±√10
Fantastic. You nailed it
(х^2+6х+8)(х^2+6х)=9
x(x+2)(x+4)(x+6)=9
(x²+6x)(x²+6x+8)=9
Let y=x²+6x
y(y+8)=9
y²+8y=9
y²+8y-9=0
(y+9)(y-1)=0
y-1=0
x²+6x-1=0
x²+6x+9=10
(x+3)²=10
|x+3|=√10
x+3=±√10
x=-3±√10 ❤❤
y+9=0
x²+6x+9=0
(x+3)²=0
x+3=0
x=-3; multiplicity of 2 ❤️ ❤
Fantabulous. You are really on point, man.
X^2+6X=t
-3
Let x+3 = y
(y-3)(y-1)(y+1)(y+3)=9
(y^2-9)(y^2-1)-9=0
y^4-9y*2-y*2+9-9=0
y*4-10y*2=0
y*2(y*2-10)=0
y = 0 y= sqrt(10) y= -sqrt(10)
x= -3 x= -3+sqrt(10) x= -3-sqrt(10)
Excellent. You nailed it, man.
一3 in 20 seconds
x^2+6x+4=a buradan x^2+6x=a-4
(a-4)(a+4)=9
a^2-16=9
a^2=9+16=25 , a=5, a=-5
x^2+6x+4=5 x^2+6x+4=-5
x^2+6x-1=0 x^2+6x+9=0
x=-3+-*(-3)^2-(-1) (x+3)^2=0
x=-3+-*9+1 x+3=0
x=-3+-*10 x=-3
x=-3-*10 x=-3+*10
Təşəkkürlər.
Nice one! 👏
3?
This is a quartic equation. Shouldn’t there be 4 solutions for x?
the root x=-3 is a double root, plus x=-3(+/-)sqrt(10) adds up to 4 roots.
jbut
simplest solution is to symmetrize. Define x=y-3. one line gives y^2=0 (double root), and y^2=10. Done.
A Chinese 8th grader can solve it in 5 minutes
Indian 6th grader can solve this in 6 minutes
@@PP-cu3fvso can an American 8th grader lol, this is very low level math