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Home  >>  TN XII Math  >>  Complex Numbers
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Solve : $6x^{4}-25x^{3}+32x^{2}+3x-10=0$ given that one of the roots is $2-i$.

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Toolbox:
  • In a polynomial equation with real coefficients,complex roots occur in conjugate pairs.
Step 1:
$6x^4-25x^3+32x^2+3x-10=0$
One of the roots of this equation is $2-i$
Therefore $2+i$ is also a root.
Their sum=4.
Their product=4+1=5
The corresponding quadratic factor is $x^2-4x+5$
Step 2:
Let $ax^2+bx+c$ be the other quadratic factor.
Comparing the coefficient of $x^4$,
We have $a=6$
Comparing the constants,
We have $5c=-10 \Rightarrow c=-2$
Step 3:
The other factor is $(6x^2+bx-2)$
Comparing the coefficient of $x$,
$3=5b+8$
$\Rightarrow b=-1$
Step 4:
The other two roots are obtained by solving
$6x^2-x-2=0$
$6x^2-4x+3x-2=0$
$2x(3x-2)+(3x-2)=0$
$(2x+1)(3x-2)=0$
$\Rightarrow x=\large\frac{-1}{2},\frac{2}{3}$
The roots are $2\pm i,\large\frac{-1}{2},\frac{2}{3}$
answered Jun 11, 2013 by sreemathi.v
 

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