One root of x^{2}+px+q=0, for p,q real, is x=2+3i. How

abiejose55d

abiejose55d

Answered question

2022-04-24

One root of x2+px+q=0, for p,q real, is x=2+3i. How do you find p and q?

Answer & Explanation

otposlati9u8

otposlati9u8

Beginner2022-04-25Added 19 answers

Step 1
Note that if α and β are the roots, then we have
(xα)(xβ)=0
x2(α+β)x+αβ=0
Since your equation is monic,
p=-(2+3i+2-3i)=-4

RormFrure6h1

RormFrure6h1

Beginner2022-04-26Added 13 answers

Step 1
The roots must occur in conjugate pairs (why?), so the polynomial must actually be
(x23i)(x2+3i)
=x24x+13
because
(x-z)(x-z¯)=x2-2Rez+|z|2

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