Polynominals. (1 Viewer)

[12o9]

Um.. Just a few polynominal questions I can't do. Hope you guys could help . Thanks in Advance.

1. The polynominal (a-2)x²+(b+7)x-(c-8) has 3 distinct zeroes. What are the values of a,b,c?

2. Show that the equation 2x³ -3x²-7=0 has at least one root in the interval 2< x <3

3. Given px²+qx+1=0 has a root x=1
a) Show that qx²+px+1 also has a root at x=1
b) Write down the other roots of each equation and prove that their sum is the opposite of their product.

For 3, I've done part a, but I'm not so sure what part b means by 'prove that their sum is the opposite of their product'.

By the way, are we required to know 'Newton's Method of Approximation' because I've encountered a few questions which require it.

Thanks in Advance.

 

[tommykins]

回复: Polynominals.

1. Can't have 3 distinct zeros because it's to the power of 2.
2. sub in x = 3, then sub in x = 2. change in sign means there is a root between these intervals.

Too tired to be thinking for 3.

 

[12o9]

Re: 回复: Polynominals.

1. Can't have 3 distinct zeros because it's to the power of 2.
2. sub in x = 3, then sub in x = 2. change in sign means there is a root between these intervals.

Too tired to be thinking for 3.

Ah. I think the question meant 2 distinct zeros =/.

For 2, I'm not so sure what you mean by 'change in sign' sorry but could you explain it again? Don't worry if you're too tired though .

Thanks .

 

[lyounamu]

Re: 回复: Polynominals.

Ah. I think the question meant 2 distinct zeros =/.

For 2, I'm not so sure what you mean by 'change in sign' sorry but could you explain it again? Don't worry if you're too tired though .

Thanks .

Root is where the equation becomes 0. So if the root is to be there, if you sub x=2 and x=2, one of them will be positive and the another one will be negative meaning that THERE IS 0 between them .

So f(2) = -3 and f(3) = 20 therefore f(x) = 0 must be between the two. Therefore, the root lies between x=2 and x=3.

3. Given px2+qx+1=0 has a root x=1
a) Show that qx2+px+1 also has a root at x=1
b) Write down the other roots of each equation and prove that their sum is the opposite of their product.

a) f(x) = px^2 + qx + 1
f(1) = p + q + 1 = 0 as the root is at x=1

g(x) = qx^2 + px + 1
g(1) = q +p + 1 which is 0 therefore, the root is at x=1

b) px^2 + qx + 1 = 0 has its root at x=1 which means it is divisible by (x-1)
Using the long division method, you get: f(x) = (x-1)(px+p+q) + (1+p+q) = (x-1)(px+p+q) + 0 since (1+p+q) = 0
So f(x) = (x-1)(px+p+q) and that is also equal to (x-1)(px-1) since p+q = -1
So the roots are x=1 or 1/p

Similarly g(x) = (x-1)(qx+p+q) which is also equal to (x-1)(qx-1)
so the roots are x=1 or 1/q

By "opposite", I think they are referring to the sign being opposite because they must be making a connection between the questions.

So the sum of roots for the f(x) is 1+1/p. If we somehow take p and q being negative, it makes sense as
1+1/p = positive and 1/p = negative.
AND 1+1/q = positive and 1/q = negative.
You may argue that p = fraction meaning that my assumption is wrong but I don't think that's the case of this question as p and q must be the WHOLE NUMBERS looking at that function.

 

[Iruka]

I think the question is correct - if a polynomial of order 2 has 3 distinct zeros then it must be the zero polynomial, hence all coefficients are zero.

 

[tommykins]

I think the question is correct - if a polynomial of order 2 has 3 distinct zeros then it must be the zero polynomial, hence all coefficients are zero.

Haha, completely and totally forgot about that hey.

 

[12o9]

I think the question is correct - if a polynomial of order 2 has 3 distinct zeros then it must be the zero polynomial, hence all coefficients are zero.

oh hey. x).

Thanks tommykins,3unitz, lyounamu and Iruka