Polynomial and other thingss (1 Viewer)

[coyazayo]

Hi everyone

Q1. When a polynomial P(x) is divided by x^2-5 the remainder is x+4. Find the remainder when P(x)+P(-x) is divided by x^2-5

Q2. Consider the equations of the straight line and the square root,

y=2+ax and y=\sqrt{ 5+x}

Find the set of values of a, when the line intersects the graph of the square root at two distinct points. (For this question, I want to know the algebraical approach. I know that graphical method is much more quicker and easier, however I want to get to know other different ways of approaching. What are the set conditions? Thanks in advance )

 

[leehuan]

P(x) = (x^2+5)Q(x) + (x+4)
P(-x) = (x^2+5)Q(-x) + (-x+4)

P(x)+P(-x) = (x^2+5)(Q(x)+Q(-x)) + 8

Doubting my answer though, cause haven't seen these questions in a while.
________________________

I remember doing it algebraically with dy/dx and the discriminant but after a counterexample happened I'm no longer sure.

Or just equate the two and get a quadratic, then 2 solutions -> discriminant > 0

Either one of the above has failed me.

 

[Paradoxica]

I will do this for a more general form of the question

Let P(x) be an arbitrary polynomial.

Let Q(x) be the non-remainder when P(x) is divided by x²-a²

Q(x)(x²-a²) + mx + n = P(x)

Now we replace x with -x to find P(-x)

Q(-x)(x²-a²) - mx + n = P(-x)

Add the two together to yield

(Q(x)+Q(-x))(x²-a²) + 2n = P(x)+P(-x)

Then the conclusion follows.

You set a² = 5, m = 1, n = 4

 

[Paradoxica]

For your second question

Note that since y is a surd, it must be strictly positive

Square both sides of equation 2

y² = 5+x

Solving for y, we get

y = 2+ay² - 5a

The discriminant of y is in terms of a

1-8a+20a²

For there to be two distinct points of intersection, we must have two conditions:

1. The discriminant is positive

2. None of the solutions for y are negative.

So we have two inequalities

The discriminant is less than unity in order for there to be two positive intersection points.

The discriminant is positive.

1>1-8a+20a²

20a²-8a+1>0

The second inequality is true for all real values of a.

The first inequality is only true for 0<a<2⁄5

The conclusion follows.

 

[KingOfActing]

The only algebraic way to do the second question rigorously that I can think of is to consider the cases a < 0, a = 0, 0 < a <= 2/5, a > 2/5, and the find the solutions in every case through a bunch of inequalities - not fun. You can't use the discriminant because it's always positive anyway.

EDIT: Nevermind, Paradoxica's got you covered

 

[coyazayo]

For your second question

Note that since y is a surd, it must be strictly positive

Square both sides of equation 2

y² = 5+x

Solving for y, we get

y = 2+ay² - 5a

The discriminant of y is in terms of a

1-8a+20a²

Hi. I don't understand what you meant by unity. Could you elaborate on it? Thanks

For there to be two distinct points of intersection, we must have two conditions:

1. The discriminant is positive

2. None of the solutions for y are negative.

So we have two inequalities

The discriminant is less than unity in order for there to be two positive intersection points.

The discriminant is positive.

1>1-8a+20a²

20a²-8a+1>0

The second inequality is true for all real values of a.

The first inequality is only true for 0<a<2⁄5

The conclusion follows.

Hi could you elaborate on what you mean by unity? I dont understand why for there to be a positive solution, the inequality must be less than unity

 

[Paradoxica]

unity -> derived from unit

basically unity means 1


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[coyazayo]

unity -> derived from unit

basically unity means 1


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But how does the inequality being less than the unity make it so that it only has positive solutions?

 

[Paradoxica]

Look at the form of the solutions for y

2ay = 1 ± √δ

"a" must be positive, but that tells us nothing, so we can discard that condition

1-√δ

if δ is greater than 1, then so is √δ

but then that would mean one of the values of y is negative

and we can't have that.