Solving a Nice Rational Equation in Two Ways

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PLAYLISTS 🎵 :

[armacham]

Here is another substitution method:
define a = x^2 + x + 2
then you can sub it in at two places:

x^2 + 2x + 2 = 2x^2 / (x^2 + x + 2)
a + x = 2x^2 / a
aa + ax = 2x^2
2x^2 - ax - aa = 0

from here you can apply the quadratic formula -- even though a is not a constant. The quadratic formula works even when A is a function of x

"a" = 2
"b" = -a
"c" = -aa

x = ( a +- sqrt(aa + 2*4*aa) ) / 4

x = ( a +- sqrt(aa*9) ) / 4

x = ( a +- 3a ) / 4

which gives us two possibilities:

x = ( a + 3a ) / 4
x = ( a - 3a ) / 4



x = 4a/ 4 = a
x = ( -2a ) / 4 = -0.5a

x = a can be rewritten as x = xx + x + 2

which gives us xx + 2 = 0, obviously that has no real solutions and it's easy to find complex ones

x = -0.5a can be re-written as:
-2x = a
-2x = xx + x + 2
0 = xx + 3x + 2

you can factor that easily or apply the quadratic formula to get the only real solutions

[muhammadmahdidacosta]

Another method:
You can use polynomial division on the left to simplify. Then the 2 will cancel on both sides and x+2 will be in common. After simplifying, the equation will reduce to a factorable cubic

[moeberry]

Another amazing video from the Master. Just a reminder we must check the denominator of the rational function to see what x values make it 0 to make sure that the complex solutions that you found don’t make the expression undefined. But in this case both complex solutions are 👍

[snejpu]

I used another method. t=x^2+x+2, then t+x=2x^2/t. t^2+xt-2x^2=0. Solving for t, we get t=x or t=-2x. Substituting back, x^2+x+2=x or x^2+x+2=-2x. We get the same 4 solutions, 2 of which are real.

[naturelover]

The 2nd method was quite elegant. I loved that!

[xiangge]

A simpler method: the nominator of the right-hand side can be written as 2x^2 +2x+4 - 2x-4; so the right-hand side becomes 2-(2x+4)/(x^2+x+2); the "2" on both sides can be cancelled, the left-hand side becomes x(x+2); the right side is -2(x+2)/(x^2+x+2); we get the first solution of x=-2 right away; By cancelling (x+2) on both sides, we get a simpler equation: x = -2/(x^2+x+2); then we can get x^3+x^2+2x+2 = 0; this can further reduced to (x+1)(x^2+2)=0. From here we can get all roots.

[armacham]

I assumed we were only looking for real solutions, because that is the convention you have been following. So I can see how some people who didn't bother finding the imaginary solutions might feel like they failed to solve the problem completely or they might feel like you tricked them.

[sphakamisozondi]

You make solving these problems look soo easy!

[ThAlEdison]

how I did it, similar but slightly different from first method
x^2+2x+2=2x^2/(x^2+x+2)
Because the rhs has the same degree on the top and bottom, we can separate it into a constant+a rational function

x^2+2x+2=2-(2x+4)/(x^2+x+2)
x^2+2x=-2(x+2)/(x^2+x+2)
cross multiplying
x(x+2)(x^2+x+2)=-2(x+2)
combining common factors
(x+2)(x^3+x^2+2x+2)=0
factor by grouping
(x+2)(x+1)(x^2+2)=0

[necro]

Equation is symmetric and has even power, assuming x not equals zero then divide by x², arrange terms, make a variable change for x+2/x and done.

[InnocentNeuron]

I think there may be another relatively straight forward method which will yield all 4 solutions.
A little figuring out by assuming that the quartic may be a perfect square of the form (x^2+px+q)^2 and so on shows that:
(x^2 + 3/2 x + 2)^2 = x^4 + 3x^3 + 25/4 x^2 + 6x +4, now subtract
0 = x^4 + 3x^3 + 4x^2 + 6x +4, now take square roots on both sides, to find:
x^2 + 3/2 x + 2 = +/- 3/2 x
which yields x = {-2, -1, sqrt(2)i, -sqrt(2)i}

[misterdubity]

I did brute force at first, then for my 2nd method I substituted y = x^2 + x + 2 --> y + x = 2x^2/y then later made z = y/x --> z={1, -2} which led to same solutions.

[jayantsingh]

After cross multiplication we get eqn :
X^4 + 3x^3 + 4x^2 + 6x +4 =0
or, (x+2)(x+1)(x^2 + 2) = 0
So, x = -1 or -2 or i*2^1/2 or -i*2^1/2 .

[zthfeug]

Nice! I have another approach in mind. RHS can be:

2x^2/ (x^2+x+2) = (2x^2+2x+4)/(x^2+x+2) - (2x+4)/(x^2+x+2) = 2 - (2x+4)/(x^2+x+2)

so the original equation becomes:

x^2+2x+2 = 2 - (2x+4)/(x^2+x+2) {the 2's canncel out}

x^2+2x = -(2x+4)/(x^2+x+2)


so (x^2+2x) * (x^2+x+2) + (2x+4) = 0, you group the common terms, you end up with (x+2)(x^3+x^2+2x+2) = 0. you factor the second parenthesis and you end up with (x+2)(x+1)(x^2+2)=0, from which you can get all the answers. :)

[TedHopp]

If you're going to include complex solutions, you do have to consider that x^2+x+1 could be equal to zero. So for both methods, you need to go back and check that none of the solutions are extraneous.

[scottleung]

Got it using the first method, except I used long division instead of factoring, which I struggle with.

[fredericocatapannarciso]

My approach was to call the denominator y and the LHS y+x and then solve the quadratic for y

[davidseed]

I love the y substitution. I coukd see the two parts looking similar but didnt see the division by x . Could we do it without the division? z=x²+2,
(z+x)(z+2x)=2x²
Z²+3zx+2x²=2x²
z(z+3x)=0
So z=0 or z=-3x
Yes you could.

[MrLidless]

If you rewrite the numerator on the RHS as 2(x²+x+2)-2(x+2) to get rid of 2 on both sides, the equation can immediately be simplified to x(x+2) = -2(x+2)/(x²+x+2), which of course can be further simplified and worked on.

[syfq_]

for me i did this by long division:
x² + 2x + 2 = 2x² / (x² + x + 2)
becomes x² + 2x + 2 = 2 - (2x + 4)/(x² + x + 2)
the 2 cancel out and 2x + 4 has a common factor of 2;
x² + 2x = -2(x+2)/(x² + x + 2)
x(x+2) = -2(x+2)/(x² + x + 2)
multiply the denominator and move everything to LHS:
x(x+2)(x² + x + 2) +2(x+2) = 0
factor out x+2 and distribute the x:
(x+2)(x³ + x² + 2x + 2) = 0
for the right equation notice that x=-1 is a solution, x+1 is a factor
therefore (x+2)(x+1)(x² + 2) = 0
x=-2, -1, √2 i, -√2 i