Help plz About Complex

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wiselyX

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Given a=3+7i is root of the quadratic equation X^2+px+q=0, where p and q are real numbers.

Question a)Someone claims B=3-7i is also root of the equations.Yes or No ??Explain your answer
 
Have you learned a theorem about complex roots of polynomials with real coefficients?

If not, there are several ways you might convince yourself in the case of a quadratic. Please either make an attempt, or just tell us what you have learned that might be useful. One such thing would be the quadratic formula; another could be the factor theorem.
 
Dr.Peterson said:
Have you learned a theorem about complex roots of polynomials with real coefficients?

If not, there are several ways you might convince yourself in the case of a quadratic. Please either make an attempt, or just tell us what you have learned that might be useful. One such thing would be the quadratic formula; another could be the factor theorem.
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Is it complex conjugate?? But it is not in my syllabus, so have other ways to solve this problem
 
wiselyX said:
Given a=3+7i is root of the quadratic equation X^2+px+q=0, where p and q are real numbers.
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Notation: suppose \(z=a+bi\) then the congregate is \(\overline{~z~}=a-bi\).
Can you show that \((x-z)(x-\overline{~z~})=x^2-2ax+a^2+b^2~?\)
 
pka said:
Notation: suppose \(z=a+bi\) then the congregate is \(\overline{~z~}=a-bi\).
Can you show that \((x-z)(x-\overline{~z~})=x^2-2ax+a^2+b^2~?\)
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Sorry, I haven't learned complex conjugate, and I think my math teacher won't let me use this way, so have other ways to solve?Plz
 
wiselyX said:
Given a=3+7i is root of the quadratic equation X^2+px+q=0, where p and q are real numbers.

Question a)Someone claims B=3-7i is also root of the equations.Yes or No ??Explain your answer
Click to expand...
One of the ways to do this, w/o using the term (complex conjugate):

Since x = 3 + 7i is a root of the equation, we have:

(3 + 7i)^2 + p*(3+7i) + q =0

You should be able to get two expressions (collecting real and imaginary parts)

one describing 'p' and​
​
second describing 'q''.​

Now, using those two expressions for 'p' and 'q' evaluate:

(3-7i)^2 + p*(3-7i) + q = ?
 
wiselyX said:
Given a=3+7i is root of the quadratic equation X^2+px+q=0, where p and q are real numbers.

Question a)Someone claims B=3-7i is also root of the equations.Yes or No ??Explain your answer
Click to expand...
Given that 3+ 7i is a root of x^2+ px+q we must have (3+ 7i)^2+ p(3+ 7i)+ q= 9- 49+ 42i+ 3p+ 7p i+ q= (3p+ q- 40)+ (7p+ 42)i= 0. The real and imaginary parts must each be 0 so we have the two equations 3p+ q- 40= 0 and 7p+ 42= 0. Solve the second equation for p and then put that value of p into the first equation to get an equation you can solve for q.

Once you know p and q, put 3- 7i for x in that equation and show that x^2+ px+ q= 0.
 
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A quadratic with real coefficients have complex root iff the discriminant in the quadratic formula is negative.

Now look at the quadratic formula. x = [math] \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]
Now assume the discriminant is negative so the sqrt of the discriminant is an imaginary number, let's call it S

So x =[math]\dfrac{-b \pm S}{2a}[/math]What does this tell you about complex roots??
 
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