[UPLIFTJEE]

If for $\mathrm{z}=\alpha+\mathrm{i} \beta,|\mathrm{z}+2|=\mathrm{z}+4(1+\mathrm{i})$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation [JEE (Main) 08 April 2023 S I]
(A) $\mathrm{x}^{2}+7 \mathrm{x}+12=0$
(B) $x^{2}+2 x-3=0$
(C) $x^{2}+x-12=0$
(D) $x^{2}+3 x-4=0$

Log in to download PDF

[UPLIFTJEE]

Solution

$\mathrm{z}=\alpha+\mathrm{i} \beta \Rightarrow|\mathrm{z}+2|=\mathrm{z}+4(\mathrm{l}+\mathrm{i}) \Rightarrow \sqrt{(\alpha+2)^{2}+\beta^{2}}=(\alpha+4)+\mathrm{i}(\beta+4)$

\[ \begin{aligned} & \sqrt{(\alpha+2)^{2}+\beta^{2}}=\alpha+4 \quad \ldots(\mathrm{i}), \beta+4=0 \Rightarrow \beta=-4 \\ & \text { (i) } \Rightarrow \sqrt{(\alpha+2)^{2}+16}=\alpha+4 \Rightarrow \alpha^{2}+4 \alpha+20=\alpha^{2}+8 \alpha+16 \Rightarrow \alpha=1 \end{aligned} \]

$\alpha+\beta=-3, \alpha \beta=-4$ Equation with roots -3 and -4 is $\mathrm{x}^{2}+7 \mathrm{x}+12=0$

[UPLIFTJEE]

Answer

A

[Author]

Posts