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Roots of $x^{2}+\left ( 2-i \right )x-2i=0$ are
Option: 1
$i,2$

Option: 2
$i,1$

Option: 3
$i,-2$

Option: 4
$2i,-1$

Answers (1)

best_answer

$x^{2}+\left ( 2-i \right )x-2i=0\\$

$a=1,b=2-i,c=-2i\\$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\$

$=\frac{-\left ( 2-i \right )\pm \sqrt{\left ( 2-i \right )^{2}-4\cdot 1\left ( -2i \right )} }{2}\\$

$=\frac{\left ( i-2 \right )\pm \sqrt{4+ i ^{2}-4i+8i} }{2}\\$

$=\frac{\left ( i-2 \right )\pm \sqrt{\left ( 2+i \right )^{2}}}{2}\\$

$=\frac{\left ( i-2 \right )+\left ( 2+i \right )}{2},\frac{\left ( i-2 \right )-\left ( 2+i \right )}{2}$

$= i,-2$

Alternative Method

$x^{2}+\left ( 2-i \right )x-2i=0\\$

$\Rightarrow x^{2}+2x-ix-2i=0\\$

$\Rightarrow x\left ( x+2 \right )-i\left ( x+2 \right )= 0\\$

$\Rightarrow \left ( x-i \right )\left ( x+2 \right )=0\\$

$\Rightarrow \left ( x-i \right )=0\: \: or \:\: \left ( x+2 \right )=0$

$\Rightarrow x=i\: \: or\: \: x=-2$

Posted by

Gunjita

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