Question
Solve the equation by the method of your choice.
2 x squared equals negative 12 x minus 7
Question content area bottom
Part 1
A.StartSet StartFraction negative 12 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 12 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 12 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 12 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
B.StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
C.StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 4 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 4 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 4 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 4 EndFraction EndSet nbsp
D.StartSet StartFraction negative 6 minus StartRoot 2 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 2 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 2 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 2 EndRoot Over 2 EndFraction EndSet nbsp
2 x squared equals negative 12 x minus 7
Question content area bottom
Part 1
A.StartSet StartFraction negative 12 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 12 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 12 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 12 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
B.StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 2 EndFraction EndSet nbsp
C.StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 4 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 4 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 22 EndRoot Over 4 EndFraction comma StartFraction negative 6 plus StartRoot 22 EndRoot Over 4 EndFraction EndSet nbsp
D.StartSet StartFraction negative 6 minus StartRoot 2 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 2 EndRoot Over 2 EndFraction EndSet nbsp
StartSet StartFraction negative 6 minus StartRoot 2 EndRoot Over 2 EndFraction comma StartFraction negative 6 plus StartRoot 2 EndRoot Over 2 EndFraction EndSet nbsp
Answers
GPT-4o mini
To solve the equation \( 2x^2 = -12x - 7 \), we first rearrange it into standard quadratic form:
\[
2x^2 + 12x + 7 = 0
\]
Next, we can use the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, the coefficients are:
- \( a = 2 \)
- \( b = 12 \)
- \( c = 7 \)
Now, we will calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 = 12^2 = 144
\]
\[
4ac = 4 \cdot 2 \cdot 7 = 56
\]
\[
b^2 - 4ac = 144 - 56 = 88
\]
Now we can compute the square root of the discriminant:
\[
\sqrt{88} = \sqrt{4 \cdot 22} = 2\sqrt{22}
\]
Now substituting these values into the quadratic formula:
\[
x = \frac{-12 \pm 2\sqrt{22}}{2 \cdot 2}
\]
\[
x = \frac{-12 \pm 2\sqrt{22}}{4}
\]
\[
x = \frac{-12}{4} \pm \frac{2\sqrt{22}}{4}
\]
\[
x = -3 \pm \frac{\sqrt{22}}{2}
\]
So the solutions can be expressed as:
\[
x = -3 - \frac{\sqrt{22}}{2} \quad \text{and} \quad x = -3 + \frac{\sqrt{22}}{2}
\]
We note that \( -3 \) can also be rewritten as \( -\frac{6}{2} \), resulting in:
\[
x = \frac{-6 - \sqrt{22}}{2} \quad \text{and} \quad x = \frac{-6 + \sqrt{22}}{2}
\]
Thus, the final solution set is:
\[
\left\{ \frac{-6 - \sqrt{22}}{2}, \, \frac{-6 + \sqrt{22}}{2} \right\}
\]
Looking at the options provided in your question:
- **Option B:** \(\left\{ \frac{-6 - \sqrt{22}}{2}, \frac{-6 + \sqrt{22}}{2} \right\}\)
**Answer:** The correct choice is Option B. \( \left\{ \frac{-6 - \sqrt{22}}{2}, \frac{-6 + \sqrt{22}}{2} \right\} \)
\[
2x^2 + 12x + 7 = 0
\]
Next, we can use the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, the coefficients are:
- \( a = 2 \)
- \( b = 12 \)
- \( c = 7 \)
Now, we will calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 = 12^2 = 144
\]
\[
4ac = 4 \cdot 2 \cdot 7 = 56
\]
\[
b^2 - 4ac = 144 - 56 = 88
\]
Now we can compute the square root of the discriminant:
\[
\sqrt{88} = \sqrt{4 \cdot 22} = 2\sqrt{22}
\]
Now substituting these values into the quadratic formula:
\[
x = \frac{-12 \pm 2\sqrt{22}}{2 \cdot 2}
\]
\[
x = \frac{-12 \pm 2\sqrt{22}}{4}
\]
\[
x = \frac{-12}{4} \pm \frac{2\sqrt{22}}{4}
\]
\[
x = -3 \pm \frac{\sqrt{22}}{2}
\]
So the solutions can be expressed as:
\[
x = -3 - \frac{\sqrt{22}}{2} \quad \text{and} \quad x = -3 + \frac{\sqrt{22}}{2}
\]
We note that \( -3 \) can also be rewritten as \( -\frac{6}{2} \), resulting in:
\[
x = \frac{-6 - \sqrt{22}}{2} \quad \text{and} \quad x = \frac{-6 + \sqrt{22}}{2}
\]
Thus, the final solution set is:
\[
\left\{ \frac{-6 - \sqrt{22}}{2}, \, \frac{-6 + \sqrt{22}}{2} \right\}
\]
Looking at the options provided in your question:
- **Option B:** \(\left\{ \frac{-6 - \sqrt{22}}{2}, \frac{-6 + \sqrt{22}}{2} \right\}\)
**Answer:** The correct choice is Option B. \( \left\{ \frac{-6 - \sqrt{22}}{2}, \frac{-6 + \sqrt{22}}{2} \right\} \)