A Nice Radical Equation

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PLAYLISTS 🎵 :

[NadiehFan]

You are messing up a little bit at 4:14 where you have rewritten the equation

x⁴ − 14x² − x + 42 = 0

as

x⁴ = 14x² + x − 42

To solve this equation, we add 2kx² + k² to both sides, which gives

(x² + k)² = (2k + 14)x² + x + (k² − 42)

Now the left hand side is a perfect square regardless of the value of k, and the quadratic in x at the right hand side will be a perfect square as well if its discriminant is zero.

However, most of the time quartic equations from contests have nice factorizations into two quadratics with _integer_ coefficients so more often than not we _do not_ have to set up and solve a cubic in k to find a suitable value of k which will make the right hand side a perfect square.

If this quartic has a factorization into two quadratics with integer coefficients, then there must exist a value of k which makes the right hand side the square of a linear polynomial in x with an integer coefficient of x, so 2k + 14 should then be the _square of an integer_ which implies that k should then be an integer multiple of ¹⁄₂. Moreover, k² − 42 = (4k² − 168)/4 will then need to be the square of an integer multiple of ¹⁄₂.

So, we really only need to check values of k that make 2k + 14 equal to 1, 4, 9 ... and see if this makes k² − 42 the square of an integer multiple of ¹⁄₂. The first possibility is 2k + 14 = 1 which implies k = −¹³⁄₂ and with this value of k we have k² − 42 = (−¹³⁄₂)² − 42 = ¹⁶⁹⁄₄ − ¹⁶⁸⁄₄ = ¹⁄₄ = (¹⁄₂)², as required. So we select k = −¹³⁄₂ and with this value of k our quartic equation becomes

(x² − ¹³⁄₂)² = x² + x + ¹⁄₄

which gives

(x² − ¹³⁄₂)² = (x + ¹⁄₂)²
(x² − ¹³⁄₂)² − (x + ¹⁄₂)² = 0
(x² − ¹³⁄₂ + x + ¹⁄₂)(x² − ¹³⁄₂ − x − ¹⁄₂) = 0
(x² + x − 6)(x² − x − 7) = 0

and we have factored our quartic into two quadratics which are easily solved. As you can see, Ferrari beats Descartes every time when it comes to factoring a quartic.

At 7:00 you claim that the quartic

a² − 2ax² + x⁴ = a + x

would be very difficult to solve. Well, I don't think so and I'll take up the challenge. First we write this quartic in standard form as

x⁴ − 2ax² − x + a² − a = 0

There is no term with x³, so if this is to factor into two quadratics we must have

(x² + px + q₁)(x² − px + q₂) = 0

for some p, q₁, q₂. Expanding this we have

x⁴ + (q₁ + q₂ − p²)x² + p(q₂ − q₁)x + q₁q₂ = 0

and equating corresponding coefficients we have

q₁ + q₂ − p² = −2a
p(q₂ − q₁) = −1
q₁q₂ = a(a − 1)

From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1 and then it is not hard to find the solution triple (p, q₁, q₂) = (1, −(a − 1), −a) as well as (p, q₁, q₂) = (−1, −a, −(a − 1)) because inverting the sign of p will simply swap the values of q₁ and q₂ for the same factorization. With (p, q₁, q₂) = (1, −(a − 1), −a) we have

(x² + x − a + 1)(x² − x − a) = 0

so

x² + x − a + 1 = 0 ∨ x² − x − a = 0

and then it is just a matter of solving these two quadratic equations in x using the quadratic formula.

This was Descartes' method of course, but we can do the same with Ferrari. Then we rewrite the equation as

x⁴ = 2ax² + x − a² + a

and adding 2kx² + k² to both side this gives

(x² + k)² = (2k + 2a)x² + x + (k² − a² + a)

Now this looks pretty hopeless, but it is not. We want to find a value of k which makes the right hand side a square (px + q)² = p²x² + 2pqx + q² of a linear polynomial px + q, but since the coefficient 2pq of x needs to be 1, and therefore independent of a, it makes sense to try values of k which make both p² = 2k + 2a and q² = k² − a² + a independent of a. So we start by trying 2k + 2a = 1 which is true if k = ¹⁄₂ − a and this makes q² = (¹⁄₂ − a)² − a² + a = ¹⁄₄ independent of a as well, as required. So, we get

(x² + ¹⁄₂ − a)² = (x + ¹⁄₂)²
(x² + ¹⁄₂ − a)² − (x + ¹⁄₂)² = 0
(x² + ¹⁄₂ − a + x + ¹⁄₂)(x² + ¹⁄₂ − a − x − ¹⁄₂) = 0
(x² + x − a + 1)(x² − x − a) = 0

which is of course the same factorization as obtained with Descartes' method.

At 11:02 you emit surprise that only one solution of the quartic equation is a solution of the original equation. But this is really not surprising at all considering that the quartic equation resulted from squaring the original equation twice to eliminate both square roots. If we solve the quartic equation starting from the factored form

(x² + x − 6)(x² − x − 7) = 0

we have

x² + x − 6 = 0 ⋁ x² − x − 7 = 0
(x − 2)(x + 3) = 0 ⋁ (x − ¹⁄₂)² = ²⁹⁄₄
x = 2 ⋁ x = −3 ⋁ x = ¹⁄₂ + ¹⁄₂√29 ⋁ x = ¹⁄₂ − ¹⁄₂√29

and indeed each of these solutions is the unique solution of _one_ of the equations

√(7 − √(7 + x)) = x
√(7 + √(7 + x)) = −x
√(7 + √(7 + x)) = x
√(7 − √(7 + x)) = −x

[Aman-simple]

Last one surprise me...
Just how beautiful ❤️

[WahranRai]

Plagiat, copy past not allowed !
1:35 x=2 is an obvious solution ( 16 - 56 - 2 + 42 = 0).
Polynomial division ---> P(x) = x^4 -14*x^2 -x + 42 = (x-2)*(x^3+2*x^2 -10x -21) --->
x = -3 is an obvious solution of the cubic ---> P(x) = (x-2)*(x+3)*(x^2-x-7) etc ...

[Hobbitangle]

Quite an obvious substitution
y = √(7+x) > 0
results into the system of equations
7-y=x²
7+x=y²

Subtracting the first equation from the second one:
x+y=y²-x²=(y-x)(x+y)
what can be easily factorized
(x+y)(x-y+1)=0

The first case
x+y=0, x=-y
actually has no solutions because both
x>0, y>0
The second case
x-y+1=0
y=x+1
Substituting into the first equation
7-(x+1)=x²
x²+x-6=0
x={-3;2}
The only positive root shoul be selected, because x > 0

The only root as an answer:
x=2

[moeberry]

If you look at the left side function it’s domain is -7 to 42 inclusive and is always decreasing while the function x is always increasing and by inspection x=2 is a solution so it must be the only solution also I should say they are both continuous functions.

[tixanthrope]

inspection yields 2 is a sol'n
and -3 satisfies the 4 degree polynomial
then the remaining polynomial is x² - x - 7

[coreyyanofsky]

i wrote it as
7 = x² + √(x + 7)
and then my brain told me what the answer was before i could try anything else

[jay-]

Your accent makes the maths that much more enjoyable.. where are you from?

[giuseppemalaguti]

x=2(ovviamente)....elevo al quadrato risulta x^4-14x^2-x+42=0..risolvo la quartica in

[fedorhotabich]

left is decreasing and right is increasing functions and positive --> there is only 1 real root it's 2

[akifbaysal]

The easiest way I can see 2 is the only solution rather than other quartic sol like -3, (1-sqrt(29))/2, (1- sqrt(29))/2: the problem from its initial setup with sqrt terms tells me that x must be greater than zero and less than sqrt(7). This condition filters all other quartic sol.s except 2.. The curve, slope, graph intersectiin arguments etc. may be nice to put forward, , but why wasting time on such arguments while some easy, obvious filter was available to us from the initial statement of problem immediately..

[Archimedes_Notes]

First you will have to find the domain. Then it becomes clear that 2 is the only solution.

[michaelpurtell]

Started with y=root(7+x) got solutions watched video and still not sure why all the other stuff

[Ssilki_V_Profile]

LHSDecreases, rhs increasing -> only 1 solution. Guess and check - 2 is a solution.

[rob]

√(7 + x) = y
√(7 - y) = x

7 + x = y^2
7 - y = x^2
x + y = y^2 - x^2 = (y - x)(y + x)
(x + y)(1 + x - y) = 0
y = -x or y = x + 1
x^2 - x - 7 = 0
x = (1 ± √29)/2
or
7 + x = x^2 + 2x + 1
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = 2