Let's Solve A Nice System

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[@NadiehFan]

At 5:14 when you have the system s² − 2p = 13 s³ − 3ps = 35 you could also multiply both sides of the first equation by 3s and both sides of the second equation by 2. Then, when you subtract the second equation from the first you immediately have s³ = 39s − 70 or s³ − 39s + 70 = 0 To factor this cubic once you know s = 2 is a root a simpler way is to subtract 2³ − 39·2 + 70 = 0 from this equation, which gives (s³ − 2³) − 39(s − 2) = 0 (s − 2)(s² + 2s + 4 − 39) = 0 (s − 2)(s² + 2s − 35) = 0

[@SyberMath]

Definitely!

[@repsur5997]

And how you continue? Also why s=2?

[@NadiehFan]

@@repsur5997 Once we have the cubic s³ − 39s + 70 = 0 the easiest way to solve this is to start by checking potential rational solutions using the rational root theorem, and then we quickly find that s = 2 is a solution, since 2³ − 39·2 + 70 = 0. The factor theorem then tells us that s − 2 is a factor of s³ − 39s + 70. If fact we can write the cubic equation as (s − 2)(s² + 2s − 35) = 0 In turn, the quadratic s² + 2s − 35 can also be factored as is shown in the video, and then we have (s − 2)(s − 5)(s + 7) = 0 so the solutions of the cubic equation are s = 2, s = 5 and s = −7. For each of these three values of s, you can the easily find the corresponding value of p by substituting each of these values of s in the equation s² − 2p = 13 So, for s = 2 we get 4 − 2p = 13 which gives p = −⁹⁄₂, for s = 5 we get 25 − 2p = 13 which gives p = 6, and for s = −7 we get 49 − 2p = 13 which gives p = 18. Therefore, the system in s and p has the solution pairs (s, p) = (2, −⁹⁄₂), (s, p) = (5, 6) and (s, p) = (−7, 18). But these solution pairs give only the sum s and the product p of x and y, so we still need to find x and y which we can do in different ways. In the video, Sybermath uses Vieta's properties of the roots of a quadratic equation which basically say that if you have a quadratic equation t² + at + b = 0 the the sum of its two solutions is −a and the product of its solutions is b. So, if, for example, we take the solution pair (s, p) = (5, 6) which tells us that x + y = 5 and xy = 6, then we know that x and y must be the solutions of the quadratic equation t² − 5t + 6 = 0. This equation easily factors as (t − 2)(t − 3) = 0 so we know the t = 2 and t = 3 are the solutions of this equation, and since the order of the solutions does not matter this gives us two solution pairs (x, y) = (2, 3) and (x, y) = (3, 2) for our original system of equations in x and y. From each of the other two solutions pairs (s, p) = (2, −⁹⁄₂) and (s, p) = (−7, 18) we can similarly get two solution pairs (x, y) each, so all in all we get _six_ solution pairs (x, y) for our original system of equations in x and y. You don't need to use Vieta's properties of the roots of a quadratic equation to solve a system like x + y = 5 xy = 6 where you need to find two numbers whose sum and product are given. There are also other approches to solve such a system of equations. An obvious method is to get y = 5 − x from the first equation and substitute this into the second equation which then becomes x(5 − x) = 6 which gives 5x − x² = 6 and so x² − 5x + 6 = 0. As you can see, we end up with the exact same quadratic equation which has the solutions x = 2 and x = 3. If x = 2, then y = 5 − x = 5 − 2 = 3 and if x = 3 the y = 5 − x = 5 − 3 = 2, so we again find the solution pairs (x, y) = (2, 3) and (x, y) = (3, 2). Yet another method used the identity (a − b)² = (a + b)² − 4ab to get (x − y)² = (x + y)² − 4xy = 5² − 4·6 = 25 − 24 = 1 Now, (x − y)² = 1 means that we either have x − y = 1 or x − y = −1. From x + y = 5 and x − y = 1 we get 2x = (x + y) + (x − y) = 5 + 1 = 6 which gives x = 3 and 2y = (x + y) − (x − y) = 5 − 1 = 4 which gives y = 2 so we have the solution pair (x, y) = (3, 2). Similarly, if we start from x + y = 5 and x − y = −1 we find the solution pair (x, y) = (2, 3). Actually we don't need to do the calculation all over again, because inverting the sign of x − y simply amounts to swapping x and y.

[@redbeard5598]

By inspection, about 20 secs, 2 and 3.

[@repsur5997]

You Cooked with the first ecuation. By the way if i want to do the first ecuation with this method 1:44 i shoukd get the same result right

[@repsur5997]

I didnt understand nothing about the seco d half of the second ec but it was cool

[@asimov2144]

Didn't get all the trick about how you found the cubic fonction in "pink". May be a video with some examples so i can practice the trick ?!

[@GlorifiedTruth]

Apparently there is no equation that cannot be contorted into subjecting it into a "Hamlet" pun...

[@scottleung9587]

I got {x,y}={2,3} right away, but wasn't aware of the other pair of solutions.

[@Qermaq]

We went from algebra to urology and back again. What a satisfying stream.

[@SyberMath]

😆

[@DonRedmond-jk6hj]

try synthetic division. it's even faster.

[@SyberMath]

Noooo 😁

[@burgazadaeczanesi5338]

with this method (s=x+y and p=x.y) can we solve x^3+y^3=35 and x^4+y^4=97 equations?

[@NadiehFan]

Essentially yes, because you can express xⁿ + yⁿ as a polynomial in the elementary symmetric polynomials x + y = s and xy = p for any positive integer n (as you can easily prove by induction). Specifically, your system then becomes s³ − 3ps = 35 s⁴ − 4ps² + 2p² = 97 But when you eliminate p from this system you end up with an equation in s of degree 6, which is 3s⁶ − 840s³ + 2619s² − 7350 = 0 and sure enough, s = 5 is an integer solution of this equation, so you can factor out (s − 5) but then you are left with s⁵ + 5s⁴ + 25s³ − 155s² + 98s + 490 = 0 which is a quintic equation with _no_ rational solutions, so good luck solving this.

[@aniruddhabhagat488]

But as you factored out the (s-5) term you can simply get the real solutions from that part and the remain quintic will give 5 non real solution. This essentially makes s= 5 and p= 6 resulting x and y being 2,3 or vice versa.

[@NadiehFan]

@@aniruddhabhagat488No, that is not correct. Any algebraic equation of _odd_ degree with real coefficients _must_ have at least one real solution, so the quintic s⁵ + 5s⁴ + 25s³ − 155s² + 98s + 490 = 0 has at least one real solution (in fact it has exactly one real solution). So, if you just go with s = 5 you only find two real solution pairs (x, y) = (2, 3) and (x, y) = (3, 2) and you are still missing two other real solution pairs. To see this, you can also enter either the system x³ + y³ = 35 x⁴ + y⁴ = 97 or the system s³ − 3ps = 35 s⁴ − 4ps² + 2p² = 97 into WolframAlpha.

[@holyshit922]

Symmetric functions Sum of powers there are Newton-Girard idenities

[@rakenzarnsworld2]

x = 2, y = 3 and vice versa.

[@dimitardimitrakov2841]

Isnt it better xcubed plus ycubed equals (x+y)(13-xy) because xsq +ysq =13

[@misterdubity3073]

@9:42 you flip flopped the products that go with each sum. (2,3) S=5, P=6