Question

Which equation has the solutions x = -3 ± √3i/2 ?

Answer 1

Answer:Answer is option C : [$x^{2}$ + 3x + 3 ] =0

Note:  None of options matches with given question.

instead of "-3" , there should be "-$\frac{3}{2}$".

Step-by-step explanation:

Note:  None of options matches with given question.

instead of "-3" , there should be "$\frac{3}{2}$".  

Here, First thing you have to observe the nature of roots.

∴ x = -$\frac{3}{2}$+$\frac{\sqrt{3}}{2}$i and x = -$\frac{3}{2}$-$\frac{\sqrt{3}}{2}$

∴ [ x+($\frac{3}{2}$-$\frac{\sqrt{3}}{2}$i) ][ x+($\frac{3}{2}$+$\frac{\sqrt{3}}{2}$i) ]=0

∴ [ $x^{2}$ + x($\frac{3}{2}$+$\frac{\sqrt{3}}{2}$i)+ x($\frac{3}{2}$-$\frac{\sqrt{3}}{2}$i) + ($\frac{3}{2}$-$\frac{\sqrt{3}}{2}$i)($\frac{3}{2}$+$\frac{\sqrt{3}}{2}$i) ]=0

∴ [$x^{2}$ + $\frac{3}{2}$x + $\frac{\sqrt{3}}{2}$ix + $\frac{3}{2}$x - $\frac{\sqrt{3}}{2}$ix + (3-$\frac{\sqrt{3}}{2}$i)(3+$\frac{\sqrt{3}}{2}$i) ] =0

∴ [$x^{2}$ + 3x + ($\frac{3}{2}$-$\frac{\sqrt{3}}{2}$i)($\frac{3}{2}$+$\frac{\sqrt{3}}{2}$i) ] =0

∴ [$x^{2}$ + 3x + $\frac{9}{4}$ - ($\frac{\sqrt{3}}{2}$i)($\frac{\sqrt{3}}{2}$i) ] =0

∴ [$x^{2}$ + 3x + $\frac{9}{4}$ - ($\frac{3}{4}$) $i^{2}$ ] =0

∴ [$x^{2}$ + 3x + $\frac{9}{4}$ + ($\frac{3}{4}$) ] =0

∴ [$x^{2}$ + 3x + $\frac{12}{4}$ ] =0  

∴ [$x^{2}$ + 3x + 3 ] =0  

Thus, Answer is option C : [$x^{2}$ + 3x + 3 ] =0