Solving 5(1-5x^2)^2=1-x

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PLAYLISTS 🎵 :

[bjornfeuerbacher]

8 minutes before the premiere... probably you will use one of your usual tricks: substitute y = 1 - 5x² and then solve the system of equations x = 1 - 5y², y = 1 - 5x²? ;)

[pietergeerkens]

I did it by method 1; but written as
(5x)⁴ - 10(5x)² + (5x) + 20) = 0
which is much easier to factor into
( (5x)² + (5x) - 5 ) ( (5x)²- (5x) - 4 ) = 0.

It took me a quite a bit longer than 8:57 though.
😉

I'm still working on consistently seeing those nice substitutions as in Method 2.

[TheNarukman]

Very nice. I love to watch your channel and found that the substitution method is a very powerful method for math, not only for integration.

[khundeejai]

I used the first method with trial and error to factor quartic into two quadratic polynomials! I prefer your second method, thanks.

[moeberry]

Another great Video. I used the quartic formula to work on my algebra.

[NadiehFan]

How about this:

Multiply both sides by 5 so we can write

(25x² − 5)² = 5 − 5x

Now the left hand side is a perfect square and we want to keep it that way and make the right hand side into a perfect square as well. How can we do that? Well, we can add 2(25x² − 5)t + t² to both sides since (25x² − 5)² + 2(25x² − 5)t + t² = (25x² − 5 + t)². After some rearranging, this gives us

(25x² − 5 + t)² = 50tx² − 5x + (t² − 10t + 5)

Now we have a perfect square on the left hand side for any value of t and we have a quadratic in x on the right hand side which will also be a perfect square if its discriminant is zero, that is, if t satisfies

25 − 4·50t(t² − 10t + 5) = 0

It is easy to see that this condition is satisfied for t = ½ since 25 − 4·50·½·¼ = 0. With t = ½ our quartic equation becomes

(25x² − ⁹⁄₂)² = 25x² − 5x + ¼

which can be rewritten as

(25x² − ⁹⁄₂)² = (5x − ½)²

or

(25x² − ⁹⁄₂)² − (5x − ½)² = 0

and using the difference of squares identity this gives us

(25x² − 5x − 4)(25x² + 5x − 5) = 0

Here we can divide both sides by 5 (this is actually the factor by which we multiplied both sides at the start) to get

(25x² − 5x − 4)(5x² + x − 1) = 0

Of course the solution is now easy, the roots are

x₁ = (1+√17)/10, x₂ = (1−√17)/10, x₃ = (−1+√21)/10, x₄ = (−1−√21)/10

[GirishManjunathMusic]

Given:
5·(1 – 5x²)² = 1 – x
To find:
x

Substituting 1 – 5x² = y:
1 – y = 5x² — ①
1 – x = 5y² — ②

Subtracting ② from ①:
x – y = 5·(x² – y²)

Bringing all terms to LHS:
(x – y) – 5·(x² – y²) = 0

Using a² – b² = (a + b)·(a – b)
(x – y) – 5·(x + y)·(x – y) = 0
(x – y)·(1 – 5·(x + y)) = 0

From here, either x = y, or 5·(x + y) = 1.
Case 1: x = y:
x = 1 – 5x²

Bringing all terms to LHS:
5x² + x – 1 = 0
x² + ⅕x – ⅕ = 0
x = –⅒ ± √(1/100 + ⅕)
x = –⅒ ± ⅒√21.

Case 2: x + y = ⅕:
x + 1 – 5x² = ⅕

Bringing all terms to RHS:
5x² – x – ⅘ = 0
x² – ⅕x – 4/25 = 0
x = ⅒ ± √(1/100 + 4/25)
x = ⅒ ± ⅒√17.

Probably have calculation errors here somewhere those numbers are disgusting.
EDIT: nvm I was right this equation is just dirty.

[j.r.]

You kind of hand-waved away the factoring step in the 1st method. That was the only part of the problem that was difficult! The other steps were rote, and the overall strategy (expand, then solve) was obvious. I guess we can handle any problem if the solution to the tricky part will just magically appear for us. :)

[andrewm]

Doesn’t the quadratic formula begin with negative b? Either I’m wrong about that, or I misread your quadratic equations. I thought each answer started with -1 plus or minus yadda yadda yadda.

[neuralwarp]

Lame joke again.
Thumbs Down from me.