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±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
Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i

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

Answer 1

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

1. Subtract 3 from both sides:
\[
10x^2 - 2x + 10 = 0
\]

2. Now, identify the coefficients for the quadratic equation \( ax^2 + bx + c = 0 \):
- \( a = 10 \)
- \( b = -2 \)
- \( c = 10 \)

3. Use the quadratic formula to find the roots:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

4. Calculate the discriminant (\( b^2 - 4ac \)):
\[
b^2 - 4ac = (-2)^2 - 4 \cdot 10 \cdot 10 = 4 - 400 = -396
\]

5. Since the discriminant is negative, the solutions will be complex numbers:
\[
x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20}
\]

6. Rewrite the square root of the negative number:
\[
\sqrt{-396} = \sqrt{396} i
\]
- Factor \( 396 \) to simplify:
\[
396 = 4 \cdot 99 = 4 \cdot 9 \cdot 11 = 36 \cdot 11 = 6^2 \cdot 11
\]
- Thus:
\[
\sqrt{396} = 6\sqrt{11}
\]

7. Substitute back into the equation:
\[
x = \frac{2 \pm 6\sqrt{11} i}{20} = \frac{1 \pm 3\sqrt{11} i}{10}
\]

Thus, the solutions are:
\[
x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i
\]

From the list of choices provided, the correct answer is:
\[
\frac{1}{10} \pm \frac{3\sqrt{11}}{10} i
\]