# Fp1 question - stuck

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

https://postimg.cc/image/r3nbcyhwr/

Hi guys . Stuck on this question . I’ve formed a quadratic from the complex roots . I got x^2 -2x+5

How am I supposed to find the other root if I don’t know what p is ?? Also if someone could explain the sum of the roots thing with alpha and beta that would be good . Not sure if that links to the question ?

Hi guys . Stuck on this question . I’ve formed a quadratic from the complex roots . I got x^2 -2x+5

How am I supposed to find the other root if I don’t know what p is ?? Also if someone could explain the sum of the roots thing with alpha and beta that would be good . Not sure if that links to the question ?

0

reply

Report

#2

(Original post by

https://postimg.cc/image/r3nbcyhwr/

Hi guys . Stuck on this question . I’ve formed a quadratic from the complex roots . I got x^2 -2x+5

How am I supposed to find the other root if I don’t know what p is ?? Also if someone could explain the sum of the roots thing with alpha and beta that would be good . Not sure if that links to the question ?

**Angels1234**)https://postimg.cc/image/r3nbcyhwr/

Hi guys . Stuck on this question . I’ve formed a quadratic from the complex roots . I got x^2 -2x+5

How am I supposed to find the other root if I don’t know what p is ?? Also if someone could explain the sum of the roots thing with alpha and beta that would be good . Not sure if that links to the question ?

0

reply

Report

#3

Expand out the roots you know by setting up a quadratic, then expand it to get rid of the imaginary parts then you should be able to find missing values by inspection

0

reply

Report

#4

Divide by 2 to make the leading coefficient 1. Then look at the constant term to see that the product of the roots is 5/2. One root is so (as the coefficients of the cubic are real), another root is . How can you now find the remaining root?

0

reply

Report

#5

**Angels1234**)

https://postimg.cc/image/r3nbcyhwr/

Hi guys . Stuck on this question . I’ve formed a quadratic from the complex roots . I got x^2 -2x+5

How am I supposed to find the other root if I don’t know what p is ?? Also if someone could explain the sum of the roots thing with alpha and beta that would be good . Not sure if that links to the question ?

0

reply

Report

#6

wagwan

Sum of the roots is -b/a

so sum of roots is 5/2

1 + 2i + 1 - 2i + otherroot = 5/2

2 + root = 5/2

so the root is 0.5

Next just expand as you usually would

Sum of the roots is -b/a

so sum of roots is 5/2

1 + 2i + 1 - 2i + otherroot = 5/2

2 + root = 5/2

so the root is 0.5

Next just expand as you usually would

1

reply

(Original post by

Divide by 2 to make the leading coefficient 1. Then look at the constant term to see that the product of the roots is 5/2. One root is so (as the coefficients of the cubic are real), another root is . How can you now find the remaining root?

**_gcx**)Divide by 2 to make the leading coefficient 1. Then look at the constant term to see that the product of the roots is 5/2. One root is so (as the coefficients of the cubic are real), another root is . How can you now find the remaining root?

I can’t remember what the formula is in terms of alpha beta and gamma. Was it alpha + beta = -gamma and alpha x beta = gamma . ??

0

reply

Report

#8

(Original post by

Oh right , so is the other root 0.5 . So does the sum of all the roots have to equal the constant in the f(x) equation (after we divide it by 2)

I can’t remember what the formula is in terms of alpha beta and gamma. Was it alpha + beta = -gamma and alpha x beta = gamma . ??

**Angels1234**)Oh right , so is the other root 0.5 . So does the sum of all the roots have to equal the constant in the f(x) equation (after we divide it by 2)

I can’t remember what the formula is in terms of alpha beta and gamma. Was it alpha + beta = -gamma and alpha x beta = gamma . ??

The coefficient of the term is , so you just plug your roots in to find .

0

reply

(Original post by

wagwan

Sum of the roots is -b/a

so sum of roots is 5/2

1 + 2i + 1 - 2i + otherroot = 5/2

2 + root = 5/2

so the root is 0.5

Next just expand as you usually would

**thotproduct**)wagwan

Sum of the roots is -b/a

so sum of roots is 5/2

1 + 2i + 1 - 2i + otherroot = 5/2

2 + root = 5/2

so the root is 0.5

Next just expand as you usually would

Do you know if this is right ?

0

reply

Report

#10

(Original post by

great, thank you does the sum of the roots being -b/a apply to quadratics and cubics as well/ Also are there any more rules i need to know regarding sum of the roots. I vaguely remember one was alpha x beta = gamma and alpha + beta = - gamma

Do you know if this is right ?

**Angels1234**)great, thank you does the sum of the roots being -b/a apply to quadratics and cubics as well/ Also are there any more rules i need to know regarding sum of the roots. I vaguely remember one was alpha x beta = gamma and alpha + beta = - gamma

Do you know if this is right ?

It really depends on the degree polynomial you're doing big man,

For quad, ax^2 + bx + c you should know that the sum is, of course -b/a, the product of the roots is c/a.

For a cubic, ax^3 + bx^2 + cx + d, roots a, b, c. sum is again -b/a, but we can also multiply the roots two at a time, ab + bc + ac, this is equal to c/a, and the product abc is -d/a

One thing you notice is that as you go on, the signs alternate, -b/a, c/a, -d/a, etc

Here, i'll give one more example

for a quartic function, ax^4 + bx^3 + cx^2 + dx + e

sum of roots is -b/a (a+b+c+d)

sum of product of roots two at a time (i.e ab + ac + ad + bc + bd + cd) is c/a

sum of product of roots THREE at a time (i.e abc + abd + bcd + acd) is -d/a

and finally, the product of all four roots- abcd, which is -e/a.

One thing to remember is that sum of roots is -b/a, and you alternate everytime from there.

0

reply

(Original post by

-b/a applies to those too.

It really depends on the degree polynomial you're doing big man,

For quad, ax^2 + bx + c you should know that the sum is, of course -b/a, the product of the roots is c/a.

For a cubic, ax^3 + bx^2 + cx + d, roots a, b, c. sum is again -b/a, but we can also multiply the roots two at a time, ab + bc + ac, this is equal to c/a, and the product abc is -d/a

One thing you notice is that as you go on, the signs alternate, -b/a, c/a, -d/a, etc

Here, i'll give one more example

for a quartic function, ax^4 + bx^3 + cx^2 + dx + e

sum of roots is -b/a (a+b+c+d)

sum of product of roots two at a time (i.e ab + ac + ad + bc + bd + cd) is c/a

sum of product of roots THREE at a time (i.e abc + abd + bcd + acd) is -d/a

and finally, the product of all four roots- abcd, which is -e/a.

One thing to remember is that sum of roots is -b/a, and you alternate everytime from there.

**thotproduct**)-b/a applies to those too.

It really depends on the degree polynomial you're doing big man,

For quad, ax^2 + bx + c you should know that the sum is, of course -b/a, the product of the roots is c/a.

For a cubic, ax^3 + bx^2 + cx + d, roots a, b, c. sum is again -b/a, but we can also multiply the roots two at a time, ab + bc + ac, this is equal to c/a, and the product abc is -d/a

One thing you notice is that as you go on, the signs alternate, -b/a, c/a, -d/a, etc

Here, i'll give one more example

for a quartic function, ax^4 + bx^3 + cx^2 + dx + e

sum of roots is -b/a (a+b+c+d)

sum of product of roots two at a time (i.e ab + ac + ad + bc + bd + cd) is c/a

sum of product of roots THREE at a time (i.e abc + abd + bcd + acd) is -d/a

and finally, the product of all four roots- abcd, which is -e/a.

One thing to remember is that sum of roots is -b/a, and you alternate everytime from there.

thanks again

0

reply

Report

#12

(Original post by

this is such a great explantion , thanks so much . Only question i have is do the signs stop fluctuating after -d/a as the next one is -e/a and this has the same sign

thanks again

**Angels1234**)this is such a great explantion , thanks so much . Only question i have is do the signs stop fluctuating after -d/a as the next one is -e/a and this has the same sign

thanks again

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top