Polynomial Questions - Please help :) (1 Viewer)

[frontin]

Could somebody please help me on the following questions and show me how they are done through working,

I'm not exactly the brightest ext. maths student Thanks heaps.


Questions

1. Given two roots of 2x^3 - kx^2 + 8 = 0 are equal, find k.

2. The product of two roots of the equation 2x^3 - kx^2 - 8x + 4 = 0 is 1. Find the value of k.

3. If x^3 - 8x^2 + kx - 12 = 0 has one root equal to the sum of the other two, find the value of k.

4. Given that the equation 2x^3 - x^2 - 13x - 6 = 0 has two roots whose product is unity, solve the equation.

Answers

1. k = 6

2. k = 1

3. k = 19

4. x = -1/2, -2 or 3

Once again, thank you for your help.

 

[sle3pe3bumz]

ah i wish i could help but I've only started polynomials and had one lesson on it so yeah .. maybe try the maths forums ? They should be of a great deal of help ..

 

[frontin]

Thanks a lot!

I might just do that

PS: You'll enjoy polynomials, their kinda fun but as I said me = not the brightest :rofl:

 

[ssglain]

I'll write down some hints so you can try the questions for yourself. The full solutions & explanations are under the spoilers. I think BOS should set up sub-forums for all preliminary subjects like they have for HSC subjects.

All these questions rely on your ability to write down expressions for the sums & products of roots and to solve them simultaneously.

1. Given two roots of 2x³ - kx² + 8 = 0 are equal, find k.
Hint: Let the roots be x = a, a, b.

Spoiler

a + a + b = -(coefficient of x²)/(coefficient of x³)
i.e. 2a + b = k/2 ...(1)

a² + ab + ab = (coefficient of x)/coefficient of x³)
i.e. a² + 2ab = 0 ...(2)

a²b = - (constant term)/(coefficient of x³)
i.e. a²b = -4 ...(3)

There are numerous ways to tackle this, but the easiest is as follows:

from (2):
a(a + 2b) = 0
a = 0 or a + 2b = 0
i.e. b = -a/2
[Can you see why a =/= 0? This can only be true if the original polynomial is in the form of mx²(x - b) = 0, where m is a real number.]

Put in b = -a/2 in (3):
a²(-a/2) = -4
-a³/2 = -4
a³ = 8
i.e. a = 2

Put a = 2 & b = -a/2 = -1 in (1):
2(2) + (-1) = k/2
3 = k/2

.: k = 6

2. The product of two roots of the equation 2x³ - kx² - 8x + 4 = 0 is 1. Find the value of k.
Hint: Let the roots be x = a, b, c & ab = 1.

Spoiler

a + b + c = k/2 ...(1)

ab + ac + bc = -4
i.e. 1 + ac + bc = -4 --> ac + bc = -5 ...(2)

abc = -2
i.e. (1)c = -2 --> c = -2 ...(3)

put (3) in (2):
(-2)a + (-2)b = -5
i.e. a + b = 5/2

put (3) & a + b = 5/2 in (1):
(5/2) + (-2) = k/2
1/2 = k/2

.: k = 1

3. If x³ - 8x² + kx - 12 = 0 has one root equal to the sum of the other two, find the value of k.
Hint: Let the roots be x = a, b, c & a + b = c.

Spoiler

a + b + c = 8
i.e. c + c = 8 --> c = 4 ...(1)

ab + ac + bc = k
i.e. ab + c(a + b) = k --> ab + c² = k ...(2)

abc = 12 ...(3)

Put (1) in (3):
ab(4) = 12
i.e. ab = 3

Put (1) & ab = 3 in (2):
(3) + (4)² = k

.: k = 19

4. Given that the equation 2x³ - x² - 13x - 6 = 0 has two roots whose product is unity, solve the equation.
Hint: In maths, unity means 1. So this question will be solved like Q2, by letting the roots be x = a, b, c & ab = 1.

Spoiler

a + b + c = 1/2 ...(1)

ab + ac + bc = -13/2
i.e. 1 + ac + bc = -13/2 --> ac + bc = -15/2 ...(2)

abc = 3
i.e. (1)c = 3 --> c = 3 ...(3)

Put (3) in (1):
a + b + 3 = 1/2
i.e. a + b = -5/2

Solve a + b = -5/2 ... (4) & ab = 1 ...(5) simultaneously:

From (4): a = -5/2 - b

In (5): (-5/2 - b)b = 1
(-5/2)b - b² = 1
-5b - 2b² = 2
2b² + 5b + 2 = 0
(2b +1)(b + 2) = 0
i.e. b = -1/2 then a(-1/2) = 1 --> a = -2
or b = -2 then a(-2) = 1 --> a = -1/2

.: Solutions are x = -2, -1/2, 3

 

[Mark576]

May I ask, what school do you go to ssglain?

 

[salco]

you can also use the discriminant can't you?

or havn't you learnt that yet?

i always cheat n use rules from other topics

... did it specify how you had to do it?

 

[ssglain]

May I ask, what school do you go to ssglain?

Burwood Girls HS.

you can also use the discriminant can't you?

I'm not sure how you'd go about this with cubic polynomials. Please enlighten us.

You can "cheat" around when given that two or more roots are equal by using the rules for roots of multiplicity.
(Ext-2 course, although I can't see why).

 

[frontin]

I'll write down some hints so you can try the questions for yourself. The full solutions & explanations are under the spoilers. I think BOS should set up sub-forums for all preliminary subjects like they have for HSC subjects.

All these questions rely on your ability to write down expressions for the sums & products of roots and to solve them simultaneously.

[/spoiler]

Thank you so so so so much ssglain! You have no idea how happy I am now!

I really appreciate your help, and I hope it didn't take you too long to type that all up.

I kind of knew what to do with the roots but got lost when doing the simultaneous equations but now I get the concept and am applying it in other questions too.

I owe you a million! Frontin.

 

[ssglain]

Thank you so so so so much ssglain! You have no idea how happy I am now!

I really appreciate your help, and I hope it didn't take you too long to type that all up.

I kind of knew what to do with the roots but got lost when doing the simultaneous equations but now I get the concept and am applying it in other questions too.

I owe you a million! Frontin.

You're welcome. I'm glad that this has been helpful for you. Don't hesitate to post a question if you need help. The great thing about this sort of threads is that everyone can learn something from it. It's a learning process for me as well because I get practise at how to structure my responses so that my examiners will clearly see my argument.

It is great that you understand the concept. This understanding is the most important thing becasue it allows you to be very flexible when faced with a variety of different question types.

Good luck with your studies!