A Nice Polynomial Equation from Turkish Math Olympiads

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[@WahranRai]

1: 58 An obvious root will be of the form p/q with p divisors of 25 and q divisors of 4 : p = (1, 5, 25) and q =(1, 2, 4) --->
obvious solution in : (1/1, 1/2, 1/4, 5/1, 5/2, 5/4, 25/1, 25/2, 25/4)
Remark : try also with both p and q negative

[@NadiehFan]

My solution, without first watching your video, was as follows. Expanding both sides of the equation and then bringing all terms from the right hand side over to the left hand side we have
4x⁴ − 36x³ + 61x² + 90x + 25 = 0
As the regulars here know by now, my preferred method for solving quartics is Ferrari's method, so I started by completing the square with respect to the quartic and cubic terms. We have 4x⁴ = (2x²)² and 36x³ is twice the product of 2x² and 9x and therefore we have (2x² − 9x)² = 4x⁴ − 36x³ + 81x² so we can write the equation as
(2x² − 9x)² − 20x² + 90x + 25 = 0
Now, ordinarily, the next step would be to create a difference of two squares from this, but take a good look at what we have here. 20x² is 10 times 2x² and 90x is 10 times 9x, so if we take out a factor 10 from these two terms we have
(2x² − 9x)² − 10(2x² − 9x) + 25 = 0
Do you see what I see? 10 = 2·5 and 25 = 5² so, yes, the committee had something nice in store for us. The left hand side is a perfect square because using the identity a² − 2ab + b² = (a − b)² with a = 2x² − 9x, b = 5 we have
(2x² − 9x − 5)² = 0
and so
2x² − 9x − 5 = 0
2x² − 10x + x − 5 = 0
2x(x − 5) + (x − 5) = 0
(2x + 1)(x − 5) = 0
x = −½ ⋁ x = 5
In retrospect, 4x⁴ = (2x²)² and 25 = 5² together were a bit of a giveaway that the quartic polynomial was actually the square of a quadratic polynomial.

[@SyberMath]

Very nice but this is how you spell it:
Do u c what I c? 😜😉

[@moeberry8226]

There’s 4 roots each answer is repeated so they both have a multiplicity of 2.

[@timeonly1401]

Fantastic! ❤
In the 2nd method, at 6:07 , you can save a good bit of writing if you set u=x²-x+1 and v=x+1 to give:
(2u + 7v)² = 56uv
4u² + 28uv + 4v² = 56uv
4u² - 28uv + 4v² = 0
(2u - 7v)² = 0
2u - 7v = 0
then back-sub u & v to solve it in x.

[@SyberMath]

Thanks for sharing!

[@NadiehFan]

@timeonly1401: Still too much work. When at 6:18 SyberMath had arrived at
(2(x² − x + 1) + 7(x + 1))² = 56(x + 1)(x² − x + 1)
it is easy to see that since 56 = 4·7·2 we have
(2(x² − x + 1) + 7(x + 1))² = 4·2(x² − x + 1)·7(x + 1)
so we have
(a + b)² = 4ab
with a = 2(x² − x + 1), b = 7(x + 1) and therefore
(a − b)² = 0
It is a pity he didn't finish his first method because that would have been quite instructive because
4x⁴ − 36x³ + 61x² + 90x + 25 = 0
can be factored into two quadratics as
(2x² − 9x − 5)(2x² − 9x − 5) = 0
but also as
(2x² − 20x + 50)(2x² + 2x + ½) = 0

[@leif1075]

​@@SyberMathhold on I don't seewhy ANYONE WOULD EVER think.to.double tbexswuared minus x plus 1so.why the hell.do that..why not do.something anyone would.sctuslly.thi k of..I see pll.fscotring the x cubed plus 1..but there's dimply.NO.REASON whatsoever tomdouble.it so.tbis seems arbitrary and contrived and not.smart..wouldntyou agree?? Why not rewrite the expression asx^2 + ×^2 + 6x - × + 1 + 8 and then work from therr..THAT'S WHAT ppl.wpipd.scruslly do..seriously wouldn't you agree..? Thanks for sharing.

[@DonEnsley-mathdrum]

Problem
(2x² +5x+9)² = 56 (x³ +1)
Expand.
4x⁴ +10 x³ +18x² +10x³ +25x² +45x+18x² +45x+81=56 (x³ +1)
Combine like terms.
4x⁴-36x³ +61 x² +90x +25 = 0
Examination of this function reveals it has rational roots at x = 5, and x = -1/2
It factors as
(x-5)4x³-16x² (x-5)-19x(x-5)-5(x-5) = 0
(x-5) (4x³-16x²-19x-5) = 0
x = -½ is a root to the cubic as
(x+ ½) 4x² -18x(x+ ½)-10(x+ ½) =0
Factor out 2(x+ ½)=(2x+1)
(2x+1)(2x²-9x -5) = 0
2x²-9x -5 = (2x+1)(x-5)
4x⁴-36x³ +61 x² +90x +25 =
[(2x+1)(x-5)]² = 0
Double Roots
x ∈ { -½, 5 }

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@yoav613]

Nice and very easy😊

[@SyberMath]

Thanks a lot 😊

[@Chrisoikmath_]

Can you remind me the rational root theorem?

[@SyberMath]

Sure.
The roots are in the form of a fraction such that the leading coefficient is divisible by the denominator of the fraction and the constant term is divisible by the numerator.

[@leif1075]

Doesnt every9ne agree there's NO REASON to multiply (×^2 -× +1) by 2 and that psrt.should be stricken and not allowed.sincd its cpmtrived and arbitrary.and out of nowhere, all due respect??

[@NadiehFan]

No. At least I don't agree. Stop posting comments which only try to second guess ideas and which only reveal you have no clue. Multiplying (x² − x + 1) by 2 is done because the square of 2(x² − x + 1) will then give a term 4x⁴ which is required since expanding the left hand side (2x² + 5x + 9)² will also give a term 4x⁴.
Of course this polynomial equation from a Math Olympiad is contrived in such a way that it can easily be solved _if_ you recognize that it can be rewritten as
(2(x² − x + 1) + 7(x + 1))² = 4·2(x² − x + 1)·7(x + 1)
so we have
(a + b)² = 4ab
with a = 2(x² − x + 1), b = 7(x + 1) and therefore
(a − b)² = 0
Alternatively, as I discuss in my main comment on this video, you can use Ferrari's method to rewrite the equation as
(2x² − 9x)² − 10(2x² − 9x) + 25 = 0
where the left hand side is the square of 2x² − 9x − 5.