# Find the real values of $\mathit{x}$ and $\mathit{y}$ for which the following equation: $\sqrt{x^{2}+3x+8} + \left ( x+4 \right )\mathit{i} = \mathit{y}\left (2 + \mathit{i} \right )$

This is the third part of the multi-part question Q4.

Toolbox:
• If $a+ib=c+id$ then $a=c$ and $b=d$
• (i.e) if $z_1=z_2$,then $Re(z_1)=Re(z_2),Im(z_1)=Im(z_2)$
• If $z_1=a+ib$, $z_2=c+id$
• $z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$
• $\mid z_1z_2\mid=\mid z_1\mid\mid z_2\mid$
Step 1:
$\sqrt{x^2+3x+8}+(x+4)i=y(2+i)$
Equating the real and imaginary parts and simplifying
$\sqrt{x^2+3x+8}=2y$
On squaring
$x^2+3x+8=4y^2$-----(a)
$x+4=y$----(b)
Substituting from (b) in (a)
$x^2+3x+8=4(x+4)^2$
$(x+4)^2=x^2+8x+16$
$x^2+3x+8=4(x^2+8x+16)$
Step 2:
On simplifying we get
$3x^2+29x+56=0$
$x=\large\frac{-29\pm\sqrt{841-672}}{6}$
$\;\;=\large\frac{-29\pm 13}{6}$$=-7,-\large\frac{8}{3} When x=-7,y=-3 When x=\large\frac{-8}{3}$$,y=\large\frac{4}{3}$
The solution is $(-7,-3),(\large\frac{-8}{3},\frac{4}{3})$
answered Jun 7, 2013