Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  TN XII Math  >>  Complex Numbers
0 votes

Solve the equation $x^{4}-4x^{3}+11x^{2}-14x+10=0$ if one root is $1+2i$.

Can you answer this question?

1 Answer

0 votes
  • In a polynomial equation with real coefficients,complex roots occur in conjugate pairs.
Step 1:
One of the roots of this equation is $1+2i$
Therefore another root is $1-2i$
Sum of the roots=2.
The corresponding quadratic factor of the LHS is $x^2-2x+5$
Step 2:
Let $x^2+px+q$ be the other quadratic factor.
Therefore $x^4-4x^3+11x^2-14x+10=(x^2-2x+5)(x^2+px+q)$
Equating the constant
$\Rightarrow q=2$
Equating the coefficient of $x$
The other quadratic factor is $x^2-2x+2$
Step 3:
The other quadratic factor is $x^2-2x+2$
The corresponding roots are obtained by solving $x^2-2x+2=0$
$\Rightarrow x=\large\frac{2\pm\sqrt{4-8}}{2}$
$\Rightarrow 1\pm i$
The roots are $1\pm 2i$,$1\pm i$
answered Jun 10, 2013 by sreemathi.v
edited Jun 11, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App