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which answer shows the conjugate pair solutions to $2x^2 - 6x + 7 = 0$?…

Question

which answer shows the conjugate pair solutions to $2x^2 - 6x + 7 = 0$? (1 point)
$\bigcirc$ $6 \pm 2\sqrt{5}i$
$\bigcirc$ $\frac{3}{2} \pm \frac{\sqrt{5}}{2}i$
$\bigcirc$ $6 \pm \frac{\sqrt{5}}{2}i$
$\bigcirc$ $\frac{3}{2} \pm \frac{\sqrt{5}}{4}i$

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(2x^2-6x + 7 = 0\), \(a = 2\), \(b=-6\), \(c = 7\).

Step2: Calculate Discriminant

Discriminant \(D=b^2-4ac=(-6)^2-4(2)(7)=36 - 56=-20\).

Step3: Apply Quadratic Formula

\(x=\frac{-(-6)\pm\sqrt{-20}}{2(2)}=\frac{6\pm\sqrt{20}i}{4}=\frac{6\pm2\sqrt{5}i}{4}=\frac{3\pm\sqrt{5}i}{2}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}i\).

Answer:

\(\frac{3}{2}\pm\frac{\sqrt{5}}{2}i\) (the second option)