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0 votes
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in Algebra by (15 points)

z2+(i-2)z+6+6i=0

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1 Answer

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by (210 points)
\( z^2+(i-2)z+6+6i \) is a quadratic equation.

\( a=1 , b=i-2 , c=6+6i \)

so we can find roots easily.

we know \( i^2=-1 \)

\( \Delta=b^2-4ac=(i-2)^2-4(1)(6+6i)=(i^2-4i+4)-24-24i=-1-4i+4-24-24i=-21-28i \)

\( \sqrt{\Delta}=\sqrt{-21-28i}=\sqrt{-7(3+4i)}=\sqrt{-7} \times \sqrt{3+4i}=\sqrt{7}i \times \sqrt{3+4i} \)

let \( \sqrt{3+4i}=\sqrt{x}+i\sqrt{y} \) and bring the sides of the equation to the power of 2. so we have:

\( 3+4i=x+2i\sqrt{x}\sqrt{y}+i^2y=(x-y)+2\sqrt{xy}i \)

comparing real part and imaginary part, we get:

\( x-y=3 \) and \( 2\sqrt{xy}=4 \rightarrow xy=4 \)

so \( x=4 \) and \( y=1 \) and \( \sqrt{3+4i}=\sqrt{x}+i\sqrt{y}=\sqrt{4}+i\sqrt{1}=2+i \)

and \( \sqrt{\Delta}=\sqrt{7}i \times \sqrt{3+4i}=\sqrt{7}i \times (2+i)=2\sqrt{7}i+i^2\sqrt{7}=-\sqrt{7}+2\sqrt{7}i \).

the roots are:

\( z_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a} \)

\( z_1=\frac{-(i-2)+\sqrt{\Delta}}{2}=\frac{-i+2-\sqrt{7}+2\sqrt{7}i}{2}=\frac{2-\sqrt{7}+(2\sqrt{7}-1)i}{2}=\frac{2-\sqrt{7}}{2}+\frac{2\sqrt{7}-1}{2}i \)

\(z_2=\frac{-(i-2)-\sqrt{\Delta}}{2}=\frac{-i+2+\sqrt{7}-2\sqrt{7}i}{2}=\frac{2+\sqrt{7}+(-2\sqrt{7}-1)i}{2}=\frac{2+\sqrt{7}}{2}-\frac{2\sqrt{7}+1}{2}i \)

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