1.) x+4+x = x(x+4)(x+2) 2.) 2(x+2) = x(x+4)(x+2, if x+2=0, then both sides are 0, thus x=-2 3.) 2 = x^2+4x => 0 = x^2+4x-2, by applying quadratic formula we get x=-2±√6 Thus x = -2, x = -2 + √6, x = -2 - √6.
1/x + 1/(x+4) = x + 2 Domain x 0, x -4 Multiply by x(x+4) x³ + 6 x² + 6x -4 = 0 Eliminate the quadratic term by replacing x with y-2. y³-6y = 0 y(y²-6) = 0 y = 0 y = ±√6 Back substitute to obtain x values x ∊{ -2, 2-√6 , 2+√6 }
A third method (from 1:32): We have already obtained the equation: x + 4 + x = (x + 2) x (x + 4) , which may be written as: 2 (x + 2) = (x + 2) x (x + 4) Therefore we get: (x + 2) (x^2 + 4x - 2) = 0 The rest is trivial and known
4 line soln: let u=x+2
Eqn is 1/(u-2)+1/(u+2)=u
2u =(u^2-4)*u -> u=0,+/-sqrt(6)
So x=-2, -2+/-sqrt(6)
1.) x+4+x = x(x+4)(x+2)
2.) 2(x+2) = x(x+4)(x+2, if x+2=0, then both sides are 0, thus x=-2
3.) 2 = x^2+4x => 0 = x^2+4x-2, by applying quadratic formula we get x=-2±√6
Thus x = -2, x = -2 + √6, x = -2 - √6.
1/x + 1/(x+4) = x + 2
Domain x 0, x -4
Multiply by x(x+4)
x³ + 6 x² + 6x -4 = 0
Eliminate the quadratic term by replacing x with y-2.
y³-6y = 0
y(y²-6) = 0
y = 0
y = ±√6
Back substitute to obtain x values
x ∊{ -2, 2-√6 , 2+√6 }
x=-2 or -2+/-√6
A third method (from 1:32):
We have already obtained the equation: x + 4 + x = (x + 2) x (x + 4) , which may be written as:
2 (x + 2) = (x + 2) x (x + 4)
Therefore we get:
(x + 2) (x^2 + 4x - 2) = 0
The rest is trivial and known
@@shmuelzehavi4940 Sorry, I mixed it up.
The rest is known from the host's work, but it is *not* "trvial." It must be worked out. Learn when to use the word "trivial."
@@robertveith6383I used this word in a slang meaning.