Solve the following quadratic equations. x^2-(3 2+2 i) x+6 2 i=0 - Vidyadip

Question

Solve the following quadratic equations.
$x^2-(3 \sqrt{ } 2+2 i) x+6 \sqrt{2} i=0$

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Answer

Given fecaation is $x^2 \quad \beta \sqrt{2}+20 x+6 \sqrt{2} i=0$ $a-1, b=[3 \sqrt{2}+2 i n c-6 \sqrt{2}$
Discriminant $-b^2$ sac
$-\left[(3 \sqrt{2}+2 m)^2-4 \times 1 \times 6 \sqrt{21}\right.$
- 1B, $12 \sqrt{2} 21+\Delta^2-24 \sqrt{2}$
- $18 \quad 1212 \cdot 4-1 \because r^2-11$
- $14-1212$
5a, the given oquation has compler rook
These roots are given by
Let $\sqrt{4-12 y^2}=-3$ - bi wheie a b e R
Squarire oc bob sidek, ve get
$
\begin{aligned}
& 14-12 \sqrt{2} 1-z^2+1 b^2+2 z^2 \\
& 14-12 \sqrt{2})=\left(a^2-b^2\right)-2 a b i \quad \ldots\left[\left(2 \hat{r}^2=-1\right)\right. \\
&
\end{aligned}
$
Equaire sed ad incgioery parts, wo cet
$
\begin{aligned}
& \text { Whn }=1 \sqrt{2}, b=\frac{-5 \sqrt{2}}{3 \sqrt{2}}=-2 \\
& \text { When a }=-3 \sqrt{2} w=-\frac{-6 \sqrt{2}}{-k \sqrt{2}}-2 \\
& \therefore \quad \sqrt{t+12 \sqrt{2} i}= \pm(3 \sqrt{2}-3) \\
& \therefore \quad x-\frac{0 x y+2 i i-(7 \sqrt{2}-2)}{2} \\
& \therefore \quad x=\frac{0,8+2 i p+12 \sqrt{2}-2)}{2} \\
& \text { ee } x=\frac{(6 \sqrt{4}-2 i j-49 \sqrt{2}-3)}{2} \\
& x-3 \sqrt{2} \text { or } x=2 \\
&
\end{aligned}
$

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