# fp1 roots of a quadratic

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#1
Where a question asks you to find the quadratic equation that has a+bi as one of its roots

how do we know the first step to solving this will be realising that

(x-(a+bi))(x-(a-bi)) ??

why do i sub the pair into the place of the constants?
0
4 years ago
#2
(Original post by timebent)
Where a question asks you to find the quadratic equation that has a+bi as one of its roots

how do we know the first step to solving this will be realising that

(x-(a+bi))(x-(a-bi)) ??

why do i sub the pair into the place of the constants?
Assuming the coefficients of the quadratic are real, if a+bi is a root, then its conjugate a-bi is also a root.
Now a root of p will come from a factor of (x-p), whatever the value of p...
0
#3
(Original post by HapaxOromenon3)
Assuming the coefficients of the quadratic are real, if a+bi is a root, then its conjugate a-bi is also a root.
Now a root of p will come from a factor of (x-p), whatever the value of p...
This bit that's it.

How do i know it's in the form (x-a)(x-b)? where a and b are the roots of the equation
0
4 years ago
#4
(Original post by timebent)
This bit that's it.

How do i know it's in the form (x-a)(x-b)? where a and b are the roots of the equation
Well if a quadratic has roots then it can be written as .
Think about it, when you solve a quadratic, you set it to 0 and then you factorise and find the values of x for when each of the factors is 0. So you know that a quadratic with roots 2 and 3 can be written as (x-2)(x-3) because if you had (x-2)(x-3)=0 you would have no problem telling me the roots.
From this it follows that if one the roots,, is then the quadratic can be written as .
If a quadratic (or any degree polynomial) with real coefficients has complex root then its complex conjugate is also a root of the quadratic (this holds true for any degree polynomial with all real coefficients).
From this we know that a-bi is also a root of the quadratic equation so we can say that so the quadratic can be written as .
But generally if any polynomial of degree n with the form
has roots , then it can be expressed in the form
.
0
#5
(Original post by B_9710)
Well if a quadratic has roots then it can be written as .
Think about it, when you solve a quadratic, you set it to 0 and then you factorise and find the values of x for when each of the factors is 0. So you know that a quadratic with roots 2 and 3 can be written as (x-2)(x-3) because if you had (x-2)(x-3)=0 you would have no problem telling me the roots.
From this it follows that if one the roots,, is then the quadratic can be written as .
If a quadratic (or any degree polynomial) with real coefficients has complex root then its complex conjugate is also a root of the quadratic (this holds true for any degree polynomial with all real coefficients).
From this we know that a-bi is also a root of the quadratic equation so we can say that so the quadratic can be written as .
OHHH!!1 thanks for that, just needed some clarification on that ^-^
I really am simple minded and forget stuff sometimes
0
4 years ago
#6
Just an easier way to find the quadratic equation.

The quadratic equation will be in the form x^2 -(a+b)x + ab...

Where a = 1 root and in this case, b = it's complex conjugate

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