Question

Solve the quadratic equation 10x2−2x+13=3
10
𝑥
2
−
2
𝑥
+
13
=
3
. Which of the following expresses its solutions in the form a±bi
𝑎
±
𝑏
𝑖

?(1 point)
Responses

110±311√10i
1
10
±
3
11
10
𝑖
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
1
10
±
101
10
𝑖
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
1
5
±
3
11
5
𝑖
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

Answer 1

To solve the quadratic equation \(10x^2 - 2x + 13 = 3\), we start by moving the \(3\) to the left side of the equation:

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

This simplifies to:

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

Next, we can apply the quadratic formula, which is given by:

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

In our case, \(a = 10\), \(b = -2\), and \(c = 10\).

First, let's calculate the discriminant \(b^2 - 4ac\):

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

Now, substituting into the discriminant:

\[
b^2 - 4ac = 4 - 400 = -396
\]

Since the discriminant is negative, this means the solutions will be complex (they will include \(i\)).

Now, we substitute back into the quadratic formula:

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

Next, we rewrite \(\sqrt{-396}\):

\[
\sqrt{-396} = \sqrt{396}i
\]

Calculating \(\sqrt{396}\), we find:

\[
396 = 4 \cdot 99 = 4 \cdot 9 \cdot 11 = 36 \cdot 11
\]
\[
\sqrt{396} = \sqrt{36 \cdot 11} = 6\sqrt{11}
\]

Now we can replace \(\sqrt{-396}\):

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

This simplifies to:

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

Therefore, the solutions can be expressed in the form \(a \pm bi\) as:

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

This corresponds to:

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

The correct response is:

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