Given the roots of a quadratic equation:
[tex]\frac{2\pm i\sqrt{3}}{2}[/tex]
Let's find the equation in standard form.
Apply the standard form of a quadratic equation:
[tex]ax^2+bx+c=0[/tex]
Now, we have:
[tex](x-\frac{2+i\sqrt{3}}{2})(x-\frac{2-i\sqrt{3}}{2})=0[/tex]
Now, let's expand the equation.
We have:
[tex]\begin{gathered} x(x-\frac{2-i\sqrt{3}}{2})-\frac{2+i\sqrt{3}}{2}(x-\frac{2-i\sqrt{3}}{2})=0 \\ \\ \frac{x^2*2-x(2-i\sqrt{3})-x(2-i\sqrt{3})}{2}+\frac{7}{4}=0 \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} \frac{x(2x-2+\sqrt{3}i-(2+\sqrt{3}i))}{2}+\frac{7}{4}=0 \\ \\ \frac{x(2x-4)}{2}+\frac{7}{4}=0 \\ \\ x(x-2)+\frac{7}{4}=0 \\ \\ x^2-2x+\frac{7}{4}=0 \end{gathered}[/tex]
Therefore, the equation in standard form is:
[tex]x^2-2x+\frac{7}{4}=0[/tex]
ANSWER:
[tex]x^{2}-2x+\frac{7}{4}=0[/tex]
