[Grade 11 Math] Is this question impossible to solve?

[u/ATosbike]

[u/ThunkAsDrinklePeep]

Where's your work?

Those three points given to you should give you a system of equations. That may be solvable or unsolvable, but that's the first step.

Had time to look at it, so I came back to summarize your work for others who hadn't tried using P(x) + 1 ≥ x, and like me was only looking at a system.

P(x) - x + 1 ≥ 0
P(2) - 2 + 1 = 1 - 2 + 1 = 0
P(4) - 4 + 1 = 3 - 4 + 1 = 0
So 2 and four are both roots, not of P(x) but of the quartic P(x) - x + 1. Since this polynomial is strictly non-negative, any roots must have even multiplicity. Since P(x), and therefore P(x) - x + 1 are degree 4, we can rewrite the latter as the product if its roots:

P(x) - x + 1 = k(x - 2)²(x - 4)²

We can solve for k using P(5): k= 14/9

Using this we can find P(6) = 941/9 ≈ 104.555

Unless someone can find an error three if us have missed, I'm inclined to believe the answer key has a flaw.

[u/ATosbike → u/ThunkAsDrinklePeep]

my phone is broken, i was already tired after trying to write this properly in paint. But if you can understand text here you go:

We know P(2)=1 and P(4)=3

Lets rewrite this : P(x) +1 >= x
to: P(x) -x +1 >= 0

If i put 2 and 3 in that equation i get 0. So, 2 and 4 are the roots of the equation. We also know the P(x) polynom is a 4 degree polynom. So:

P(x) = k(x-2)^2 * (x-4)^2 +x-1

If i calculate P(5) using this, i get k=14/9

And if i try to get P(6) from that i get an undividable number.
(Been thinking since morning at school, at subway and at my desk right now , still no answer)

[u/No_Prior_6913 → u/ATosbike]

You made a mistake of writing p(x) as that expression when in reality it's p(x)-x+1 = k(x-2)²(x-4)²

[u/rorodar]

3 values for even just 4 variables would immediately make it impossible, even with that loose little restriction. For five, yes I believe it would be impossible. I could be wrong though

[u/No_Prior_6913 → u/rorodar]

The condition p(x) +1>= x makes it solvable

[u/MeasureDoEventThing → u/No_Prior_6913]

I don't think "solvable" is the right term. Solvable refers to existence of a solution. p(x)+1 >= x gives uniqueness, not existence, of a solution.

[u/PhoenixAsh7117 → u/rorodar]

To summarize the solution that others have already gotten to:

Define a new polynomial Q(x) = P(x) - x + 1

This polynomial is given as always >= 0

We see we have 2 roots from the given points, x=2 and 4. Both roots must have multiplicity 2 since this new function cannot be negative.

Set this function Q(x)=k(x-2)^2(x-4)^2 solve for k with the given points. Use this to solve for P(6).

[u/maniaxz → u/rorodar]

No it's solvable

[u/AdjectiveNounNNNN → u/rorodar]

The "little restriction" means two of those points are double roots, leaving only one degree of freedom that the third point takes care of.

[u/maniaxz]

​Define Q(x) = P(x) - x + 1 = 0

Now it reveals that x = 2 and x = 4 are double roots, Q(x)=14/9 (x-2)² . (X-2)²

[u/Legitimate_Log_3452 → u/maniaxz]

That is very nice to find 2 roots of Q, but how do we know they are double roots? Because else the inequality wouldn’t hold? We need Q(x) >= 0

[u/No_Prior_6913 → u/Legitimate_Log_3452]

Since Q(x) is quartic and is always greater than or equal to 0 it means the roots 2 and 4 must have even multiplicity for the graph to never dip below x axis

[u/Extension-Stay3230]

Bookmarked for later

[u/ApplicationBig9830]

the answer is given in other comments so I'll focus more on how would I do this problem
first of all I will try to convince myself that this is solvable, apparently it's a rather tricky problem of which proofing that it is solvable with one distinct answer existing is not a lot easier than finding the actual answer. The first thing here I would analysis on is the inequality, we are given it's a 4th degree polynomial that is always greater or equal to x-1 so I will draw a function out, assuming it have the maximum point of inflection, and all above the line of x-1. And it would be pretty obvious that a gotta be positive, so our shape should look some what similar. Now if we look at the points, P(2)=1, and if we try to plot that on the graph we can find it sits on the line of x-1, because 2-1=1, and same for P(4)=3, and P(5) would be further above. So with those conditions, and the fact that we know it's a fourth polynomial with positive leading coefficient, we can draw a pretty close graph for it, the general shape you are seeing now should be around the same as the one I am seeing on my paper. (why can't we post image here omfg) And another thing we know is that the point (2,1) and the point (4,3) should be where our function P is tangent to x-1, which means the derivative of the function at those x values should be 1. those things will tell us 32a+12b+4c+d=256a+48b+8c+d=1, along with the 3 given equations should be enough for us to solve the system.

32a+12b+4c+d=1

256a+48b+8c+d=1

16a+8b+4c+2d+e=1

256a+64b+16c+4d+e=3

625a+125b+25c+5d+e=18

which shall give us an answer of

a = 14/9

b = -56/3

c = 728/9

d = -445/3

e = 887/9

and plug them in to find P(6) = 941/9

[u/Fit_Fortune_7692 → u/ApplicationBig9830]

Und die Lösungen zu finden läuft über das Gaußsche Eliminationsverfahren, was auch noch ein bißchen knifflig ist

[u/ApplicationBig9830 → u/Fit_Fortune_7692]

it's not that deep

[u/last_hentaibender]

Çok zor soru harbiden yav

[u/Rich_Opportunity_554]

P(x)>=x-1

P(x)-x+1>=0

P(x)-x+1 is always bigger than or equal to 0

It means there are 2 double roots

Notice that P(2)-1=0

let x=2

P(2)-2+1=P(2)-1=0

It means 2 is root of P(x)-x+1

same for 4

P(x)-x+1=a*(x-2)^2*(x-4)^2

P(5)-5+1=a*9*1

14=9a

14/9=a

P(x)-x+1=14/9*(x-2)^2*(x-4)^2

P(x)=14/9*(x-2)^2*(x-4)^2+x-1

P(6)=14/9*16*4+5=941/9

[u/_Cahalan]

Assuming that the shape of this curve is a typical 4th order equation, it should be symmetric across the y-axis. So inputting (-2) should also equal 1 and so forth. It is a bold assumption to make with the given information, but don't be afraid to check if making the coefficients of x3 and x negative changes the fit with the 3 known points.

[u/Al2718x → u/_Cahalan]

What do you mean "typical". This is true for exactly 0% of 4th order equations.

[u/_Cahalan → u/Al2718x]

Oh my mistake, haven't had a chance to look at this in detail. If the odd terms are non zero it wouldn't be the exact same as a pure x4.

[u/AdjectiveNounNNNN → u/_Cahalan]

A "typical" quartic would not be expected to have any symmetry whatsoever.

[u/Seasplash]

You would need 5 distinct points to solve a polynomial of degree 4. With only 3 points given, there are an infinite number of curves that can satisfy the equation and conditions.

[u/AdjectiveNounNNNN → u/Seasplash]

Two of those points are multiplicity 2 roots of p(x) - x + 1, so a third point determines the whole quartic.

[u/Seasplash → u/AdjectiveNounNNNN]

You are correct. It has been a while since I have taken Algebra 2 and I overlooked the P(x) + 1 >= x .