# Fp1 question - stuck

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#1
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 ?
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2 years ago
#2
(Original post by 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 ?
Try using the sum of roots = -(this) / (that) route.
0
2 years ago
#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
2 years ago
#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?
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2 years ago
#5
(Original post by 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 ?
Any polynomial can be written in the form a(x-x1)(x-x2)...(x-xn) where a is the coefficient of the highest power of the polynomial, x1,x2,...,xn are the roots of the polynomial(solutions to f(x)=0) and n is the highest power of the polynomial. Then, its a matter of using this equation for n=3 and equating coefficients with the equation for f(x) given.
0
2 years ago
#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
1
#7
(Original post by _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?
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 . ??
0
2 years ago
#8
(Original post by 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 . ??
If the leading coefficient is 1, the product is of the roots is equal to the constant term times . The sum of the roots is equal to the coefficient of the term times . (which would've been slightly easier, but I read as being the coefficient of the term. Doesn't really matter either way.) If you want to verify this labouriously you can expand .

The coefficient of the term is , so you just plug your roots in to find .
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#9
(Original post by 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
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 ?
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2 years ago
#10
(Original post by 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 ?
-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.
0
#11
(Original post by 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.
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
2 years ago
#12
(Original post by 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 Mistyped, thats e/a
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