Question
Two Solutions for All Quadratic Equations Quick Check
5 of 55 of 5 Items
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±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
Answers
GPT-4o mini
To solve the quadratic equation \(10x^2 - 2x + 13 = 3\), we first move all terms to one side of the equation to set it to zero:
\[
10x^2 - 2x + 13 - 3 = 0
\]
This simplifies to:
\[
10x^2 - 2x + 10 = 0
\]
Now, we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 10\), \(b = -2\), and \(c = 10\). Plugging in the values, we first calculate \(b^2 - 4ac\):
\[
b^2 = (-2)^2 = 4
\]
\[
4ac = 4 \cdot 10 \cdot 10 = 400
\]
\[
b^2 - 4ac = 4 - 400 = -396
\]
Since the discriminant is negative (\(-396\)), we will have complex solutions. Now substituting back into the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10}
\]
\[
x = \frac{2 \pm \sqrt{-396}}{20}
\]
To express \(\sqrt{-396}\):
\[
\sqrt{-396} = \sqrt{396}i = \sqrt{4 \cdot 99}i = 2\sqrt{99}i = 2\sqrt{9 \cdot 11}i = 6\sqrt{11}i
\]
Now substituting this back into the equation:
\[
x = \frac{2 \pm 6\sqrt{11}i}{20} = \frac{1 \pm 3\sqrt{11}i}{10}
\]
Thus, the solutions can be expressed as:
\[
x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i
\]
Now, looking at the responses you provided, the correct one is:
**Positive \(\frac{1}{10} \pm \frac{3\sqrt{11}}{10} i\)**.
This matches the format:
\[
\frac{1}{10} \pm \frac{3\sqrt{11}}{10}i
\]
So the response is:
**Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i**.
\[
10x^2 - 2x + 13 - 3 = 0
\]
This simplifies to:
\[
10x^2 - 2x + 10 = 0
\]
Now, we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 10\), \(b = -2\), and \(c = 10\). Plugging in the values, we first calculate \(b^2 - 4ac\):
\[
b^2 = (-2)^2 = 4
\]
\[
4ac = 4 \cdot 10 \cdot 10 = 400
\]
\[
b^2 - 4ac = 4 - 400 = -396
\]
Since the discriminant is negative (\(-396\)), we will have complex solutions. Now substituting back into the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10}
\]
\[
x = \frac{2 \pm \sqrt{-396}}{20}
\]
To express \(\sqrt{-396}\):
\[
\sqrt{-396} = \sqrt{396}i = \sqrt{4 \cdot 99}i = 2\sqrt{99}i = 2\sqrt{9 \cdot 11}i = 6\sqrt{11}i
\]
Now substituting this back into the equation:
\[
x = \frac{2 \pm 6\sqrt{11}i}{20} = \frac{1 \pm 3\sqrt{11}i}{10}
\]
Thus, the solutions can be expressed as:
\[
x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i
\]
Now, looking at the responses you provided, the correct one is:
**Positive \(\frac{1}{10} \pm \frac{3\sqrt{11}}{10} i\)**.
This matches the format:
\[
\frac{1}{10} \pm \frac{3\sqrt{11}}{10}i
\]
So the response is:
**Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i**.