Polynomials (1 Viewer)

[Smithereens]

Need some help:


1. Find the value(s) of k if the quadratic equation x ² -(k+2)x + k + 1 = 0 has:
a) equal roots
b) one root =5
c) consecutive roots
d) one root double the other
e) reciprocal roots.

2. Two roots of x ³ + mx ² + 15x = 7 -0 are equal and rational. Find m.

 

[ianc]

Hope this helps



Also for question 2, do the same process:
find the values for the sum and products of the roots in terms of the coefficients (in this case you will have each expression in terms of m), then substitute for the roots using the information given





edit: by the way, just noticed your hsc is in 2008. Are you just hitting the polynomials really early or are you accelerated??

 

[Kujah]

Some Extension 1 classes start polynomials as their first topic in Year 11

 

[dr baby beanie]

^^
I did and now I've forgotten how to do them. So yeah, keep revising them

 

[SoulSearcher]

^^
I did and now I've forgotten how to do them. So yeah, keep revising them

Just like any other topic

 

[ianc]

yeah the good thing about polynomials is they dont take long to pick up again....

i hate the long division though

 

[pLuvia]

yeah the good thing about polynomials is they dont take long to pick up again....

i hate the long division though

Hardly any of that though

 

[Kujah]

Got a question Tried doing the question, but all my working out is all over the place so I just gave up.

16. The product of two of the roots of x ⁴ + 2x ³ -18x - 5 = 0 is -5. Find the product of the other two roots.

 

[SoulSearcher]

Let a, b, c, d be the roots of the equation. Let ab = -5

Therefore abcd = -5
Since ab = -5
-5cd = -5
cd = 1

There's the other product.

 

[Kujah]

Oops wrong question

Suppose to be this

Solve 6x⁴ + 5x³ - 24x² - 15x + 18 =0 if the sum of two of its roots is zero.

 

[ianc]

okay thats a bit of a long one.........

first, because the sum of two of the roots is 0, let the roots be a, -a, b, c

using the sum of the roots formula:
b + c = -(5/6)
c =-(5/6) - b


Then using the product of roots formula:
-a²bc = 3
-a² = (3/bc)

Then using the sum of roots 2 at a time formula:
-a² + ab +ac - ab - ac + bc = 4
-a² + bc = 4


And then from there its just a matter of doing some algebra bashing through simulatenous equations