Polynomials (1 Viewer)

[joesmith1975]

The polynomial x^4+ax^2+bx+36 has a double root of x=2 find a, b

Thanks

 

[sasquatch]

This looks like its gonna take ages to write.

Let alpha = A
beta = B
gamma = Y
delta = D

x^4 + 0x^3 + ax^2 + bx + 36 = 0

A = B = 2 (equation has a double root)

A + B + Y + D = 0
4 + Y + D = 0
Y + D = -4

----------------------------

AB + AY + AD + BY + BD + YD = a
4 + 2Y + 2D + 2Y + 2D + YD = a
4 + 4Y + 4D + YD = a
4 + 4(Y + D) + YD = a

but Y + D = -14

4 - 16 + YD = a
[1] a = YD - 12

---------------------------------

ABY + ABD + AYD + BYD = -b
4Y + 4D + 2YD + 2YD = -b
4(Y + D) + 4YD = -b

but Y + D = -4

-16 + 4YD = -b
[2] b = 16 - 4YD

--------------------------------------

ABYD = 36
4YD = 36
YD = 9

Sub YD = 36 into [1]

a = 9 - 12
.:. a = -3

Sub 4YD = 36 into [2]

b = 16 - 36
.:.b = -20

 

[Mountain.Dew]

actually, i think theres a better way:

f(x) = x^4 + ax^2 + bx + 36

we know that since 2 is a double root, then f(2) = 0, f'(2) = 0

so, f(2) = 16 + 4a + 2b + 36 = 0

so 2a + b = -26 .....(1)

f'(x) = 4x^3 + 2ax + b = 0
f'(2) = 32 + 4a + b = 0

so 4a + b = -32....(2)

(2) - (1) ==> 2a = -32 + 26 ==> 2a = -6, a = -3 <--sub back into (1)

-6 + b = -26, b=-20

 

[sasquatch]

hahaha.. yeah that is better...Didn't even think of that! we havent really started much of polynomials. I was using 3 unit stuff.. but thats way seems so much easier... look how short it is compared to mine. Well... at least got the right answer!

 

[Mountain.Dew]

you will learn all this polynomial theory when u formally start the topic. they are really very simple --> just knowing what theory to apply to what situation.

but your way is equally valid, although a bit longer.

 

[sasquatch]

MUCH longer...

 

[joesmith1975]

we know that since 2 is a double root, then f(2) = 0, f'(2) = 0

How do we know this?

Don't worry about this i just when into my notes and and read the textbook and found the roots of multiplicity{r} has {r-1} roots for p'{x} and {r-2} roots for p''{x} and so on..

THANKS ANYWAYS

 

[sasquatch]

Generally a double root, means a turning point and a triple root a horizontal point of inflexion.

You can demonstrate these results as such:

If a is a double root of f(x) then,

f(x) = (x - a)²g(x)
f'(x) = (x - a)[2g(x) + g'(x)(x-a)]

When f(x) = 0, x = a
When f'(x) = 0, x = a

Or something like that.... Works similarily with a triple root..

 

[joesmith1975]

now solve P{x} over the real number system

 

[sasquatch]

Well that aint so bad...

P(x) = x⁴ - 3x² - 20x + 36

You know x = 2 is a double root hence,

(x-2)² is a factor of P(x).

Expanding: (x² - 4x + 4) is a factor of P(x). To find the other factors you must use polynomial division..

Using poly division you get..

P(x) = (x² - 4x + 4)(x² + 4x + 9)

When P(x) = 0

x = 2, x = [-4 +/- root(16 - 4(1)(9))] / 2

but x = 2 is the only real solution.