Question

question 5
1 pts
what is the solution set of the equation using the quadratic formula?
x² + 6x + 10 = 0
○ {-3 + i, -3 - i}
○ {-6 + 2i, -6 - 2i}
○ {-2i, -4i}
○ {-3 + 2i, -3 - 2i}

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for \( ax^2 + bx + c = 0 \) is \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). For \( x^2 + 6x + 10 = 0 \), \( a = 1 \), \( b = 6 \), \( c = 10 \).

Step2: Calculate Discriminant

Discriminant \( D=b^2 - 4ac=(6)^2 - 4(1)(10)=36 - 40=-4 \).

Step3: Substitute into Formula

\( x=\frac{-6\pm\sqrt{-4}}{2(1)}=\frac{-6\pm2i}{2}=-3\pm i \)? Wait, no, wait: \( \sqrt{-4}=2i \), so \( x=\frac{-6\pm2i}{2}=-3\pm i \)? Wait, no, wait, \( \frac{-6 + 2i}{2}=-3 + i \), \( \frac{-6 - 2i}{2}=-3 - i \)? Wait, no, wait the last option is \( \{-3 + 2i, -3 - 2i\} \). Wait, I made a mistake. Wait, \( b^2-4ac = 36 - 40=-4 \), so \( \sqrt{-4}=2i \). Then \( x=\frac{-6\pm2i}{2}=\frac{-6}{2}\pm\frac{2i}{2}=-3\pm i \)? But that's the first option. Wait, no, wait, the equation is \( x^2 + 6x + 10 = 0 \). Wait, maybe I miscalculated. Wait, \( a = 1 \), \( b = 6 \), \( c = 10 \). So \( b^2 - 4ac = 36 - 40 = -4 \). So \( \sqrt{-4}=2i \). Then \( x=\frac{-6\pm2i}{2}=-3\pm i \). But the first option is \( \{-3 + i, -3 - i\} \), the last is \( \{-3 + 2i, -3 - 2i\} \). Wait, no, wait, maybe I messed up the equation. Wait, no, the equation is \( x^2 + 6x + 10 = 0 \). Wait, let's recalculate: \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}=\frac{-6\pm\sqrt{36 - 40}}{2}=\frac{-6\pm\sqrt{-4}}{2}=\frac{-6\pm2i}{2}=-3\pm i \). So the first option is \( \{-3 + i, -3 - i\} \). Wait, but let's check the last option: if it's \( -3\pm2i \), then discriminant would be \( (2i)^2\times4 \)? No, wait, maybe I made a mistake. Wait, no, let's solve the equation by completing the square. \( x^2 + 6x + 10 = 0 \) => \( x^2 + 6x = -10 \) => \( x^2 + 6x + 9 = -10 + 9 \) => \( (x + 3)^2 = -1 \) => \( x + 3=\pm i \) => \( x=-3\pm i \). So the solution set is \( \{-3 + i, -3 - i\} \), which is the first option. Wait, but the last option has 2i. Wait, maybe the equation was \( x^2 + 6x + 13 = 0 \)? No, the problem says \( x^2 + 6x + 10 = 0 \). So according to the calculation, the solution is \( \{-3 + i, -3 - i\} \), which is option A. Wait, but let's check the options again:

Option 1: \( \{-3 + i, -3 - i\} \)

Option 2: \( \{-6 + 2i, -6 - 2i\} \)

Option 3: \( \{-2i, -4i\} \)

Option 4: \( \{-3 + 2i, -3 - 2i\} \)

So according to the quadratic formula, the solution is \( x=-3\pm i \), so option 1.

Answer:

\(\{-3 + i, -3 - i\}\) (Option A)