Question

Which answer shows the conjugate pair solutions to 2x2−6x+7=0
?(1 point)
Responses

6±2i5–√
6 plus or minus 2 i Start Root 5 End Root

32±i5√2
Start Fraction 3 over 2 End Fraction plus or minus Start Fraction i Start Root 5 End Root over 2 End Fraction

32±i5√4
Start Fraction 3 over 2 End Fraction plus or minus Start Fraction i Start Root 5 End Root over 4 End Fraction

6±i5√2

Answer 1

To find the solutions to the quadratic equation \(2x^2 - 6x + 7 = 0\), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(2x^2 - 6x + 7 = 0\), the coefficients are:
- \(a = 2\)
- \(b = -6\)
- \(c = 7\)

Now we can substitute these values into the quadratic formula:

1. Calculate the discriminant:

\[
b^2 - 4ac = (-6)^2 - 4(2)(7) = 36 - 56 = -20
\]

Since the discriminant is negative, this indicates the solutions will be complex.

2. Now substitute into the quadratic formula:

\[
x = \frac{-(-6) \pm \sqrt{-20}}{2(2)} = \frac{6 \pm \sqrt{-20}}{4}
\]

We can simplify \(\sqrt{-20}\):

\[
\sqrt{-20} = \sqrt{20} i = \sqrt{4 \cdot 5} i = 2\sqrt{5} i
\]

3. So now we substitute this back in:

\[
x = \frac{6 \pm 2\sqrt{5} i}{4} = \frac{6}{4} \pm \frac{2\sqrt{5} i}{4} = \frac{3}{2} \pm \frac{\sqrt{5} i}{2}
\]

Thus, the solutions to the quadratic equation \(2x^2 - 6x + 7 = 0\) are:

\[
\frac{3}{2} \pm \frac{\sqrt{5}}{2} i
\]

This corresponds to the answer:

**\(32 \pm i \sqrt{5} / 2\)** (Start Fraction 3 over 2 End Fraction plus or minus Start Fraction i Start Root 5 End Root over 2 End Fraction)