Finding quadratic equation when roots are a+3 and b+3

[tangaloomaflyer]

Hi,

I'm trying to find out how to do the following:

Find the quadratic equation with numerical coefficients when the roots are a+3 and b+3

I know the answer is x² - 4x +8 but I don't know how to get there.

I have the following hint:

Let a+3 = A
Let b+3 = B
Find the values for A and B when required equation is x² -(A+B)x + AB

I know the roots have to be complex, so I was thinking that a+3 and b+3 could be written as 3+i and 3-i, but if I put these roots in the equation I can't re-arrange it to get x² - 4x +8

Help!!

 

[pka]

Find the quadratic equation with numerical coefficients when the roots are a+3 and b+3
I know the answer is x² - 4x +8 but I don't know how to get there.!

How do you know that?

If \(\displaystyle a=0~\&~b=7\) then \(\displaystyle \large x^2-13x+30=0\) also works.
Any pair \(\displaystyle (a,b)\) will work. There is no unique answer.

 

[Ishuda]

Hi,

I'm trying to find out how to do the following:

Find the quadratic equation with numerical coefficients when the roots are a+3 and b+3

I know the answer is x² - 4x +8 but I don't know how to get there.

I have the following hint:

Let a+3 = A
Let b+3 = B
Find the values for A and B when required equation is x² -(A+B)x + AB

I know the roots have to be complex, so I was thinking that a+3 and b+3 could be written as 3+i and 3-i, but if I put these roots in the equation I can't re-arrange it to get x² - 4x +8

Help!!

With which numerical coefficients? If you know the coefficients for the linear (x) term is -4 and the constant term is 8, then take your two sets of equations and solve for A (or B) then B (or A). That is
(1) A + B = 4
(2) A B = 8
So (1) gives us
B = 4 - A
and using (2) we have
A (4 - A ) = 8
...

 

[tangaloomaflyer]

OK, full question from textbook, with answers:

Show that x = 2 is a root of equation x³+x-10 = 0. Given that the other roots p and q show that p+q = -2 and find value of pq [note p and q should be written as alpha and beta but I have changed the letter to avoid confusion with quadratic coefficients]

re-arrange: (x-2)(ax²+bx+c) = 0,

therefore
x= 2 is a root: (2-2)(a2²+b2+c) = 0,

Using x2-(p+q)x + pq = 0
p+q = -b/a
pq = c/a

then if p+q = -b/a = -2 then b = 2

(x-2)(x² +2x+5) = x³+2x²+5x-2x²-4x-10 = x³+x-10

c = 5
c/a = pq = 5


The next part was my first question, but i'll re-write using p and q as the roots:

Find the equation with numerical coefficients whose roots are p+3 and q+3

It's a guided question and the hint is:

Let p+3 be A
Let q+3 be B

Find A+B and AB when required equation is
x² -(A+B)x + AB = 0


Thanks

 

[tangaloomaflyer]

OK, so I figured it out

From the first part of the question,

p+q = -2
pq = 5

so solve p and q

q = -2-p
p(-2-p) = 5

p²+2p+5 = 0

Using quadratic formula, p = -1+2i or -1-2i,
therefore p and q are -1+2i and -1-2i (conjugate complex roots)

Then, let p+3 be A, let q+3 be B
A = 2+2i
B = 2-2i

A+B = 4+2i-2i = 4
AB = (2+2i)(2-2i) = -4i²+2i-2i+4 = 8

x²-(A+B)x+AB = 0

x² -4x +8 = 0


The weird thing is that the textbook hasn't got up to imaginary numbers yet. Is there a simpler way of doing it?