Question

Solve the quadratic equation 10x2−2x+13=3 . Which of the following expresses its solutions in the form a±bi ?(1 point)
Responses

110±101√10i
Start Fraction 1 over 10 End Fraction plus or minus Start Fraction Start Root 101 End Root over 10 End Fraction i

15±311√5i
Start Fraction 1 over 5 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 5 End Fraction i

−110±311√10i
negative Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i

110±311√10i

Answer 1

To solve the quadratic equation \( 10x^2 - 2x + 13 = 3 \), we first rearrange it into standard form:

\[
10x^2 - 2x + 13 - 3 = 0
\]

This simplifies to:

\[
10x^2 - 2x + 10 = 0
\]

Next, we can solve this quadratic equation using the quadratic formula:

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

Here, \( a = 10 \), \( b = -2 \), and \( c = 10 \). Plugging in these values:

1. Calculate \( b^2 - 4ac \):

\[
b^2 = (-2)^2 = 4
\]
\[
4ac = 4 \cdot 10 \cdot 10 = 400
\]
\[
b^2 - 4ac = 4 - 400 = -396
\]

2. Now plug this into the quadratic formula:

\[
x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20}
\]

3. Simplify \( \sqrt{-396} \):

\[
\sqrt{-396} = \sqrt{396} i = \sqrt{4 \cdot 99} i = 2\sqrt{99} i = 2\sqrt{9 \cdot 11} i = 6\sqrt{11} i
\]

4. Substituting back gives:

\[
x = \frac{2 \pm 6\sqrt{11} i}{20} = \frac{2}{20} \pm \frac{6\sqrt{11} i}{20} = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i
\]

Thus, the solutions of the equation are:

\[
x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i
\]

Looking at the provided options, we can see that the correct expression for the solutions in the form \( a \pm bi \) is:

\[
\text{Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i}
\]

Therefore, this is the answer.