# fp1 roots of a quadratic

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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?

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?

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(Original post by

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?

**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?

Now a root of p will come from a factor of (x-p), whatever the value of p...

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(Original post by

Assuming the coefficients of the quadratic are real, if a+bi is a root, then its conjugate a-bi is also a root.

**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...**How do i know it's in the form (x-a)(x-b)? where a and b are the roots of the equation

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#4

(Original post by

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

**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

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

.

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(Original post by

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

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 .

**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 .

I really am simple minded and forget stuff sometimes

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#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

Posted from TSR Mobile

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

Posted from TSR Mobile

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