Question

Find the two square roots of negative 36 plus 36 Start Root 3 End Root i.
(1 point)
Responses

3 Start Root 2 End Root minus 3 Start Root 6 End Root i, negative 3 Start Root 2 End Root plus 3 Start Root 6 End Root i
Image with alt text: 3 Start Root 2 End Root minus 3 Start Root 6 End Root i , Image with alt text: negative 3 Start Root 2 End Root plus 3 Start Root 6 End Root i

negative 3 Start Root 2 End Root plus 3 Start Root 6 End Root i, negative 3 Start Root 2 End Root minus 3 Start Root 6 End Root i
Image with alt text: negative 3 Start Root 2 End Root plus 3 Start Root 6 End Root i , Image with alt text: negative 3 Start Root 2 End Root minus 3 Start Root 6 End Root i

3 Start Root 2 End Root plus 3 Start Root 6 End Root i, 3 Start Root 2 End Root minus 3 Start Root 6 End Root i
Image with alt text: 3 Start Root 2 End Root plus 3 Start Root 6 End Root i , Image with alt text: 3 Start Root 2 End Root minus 3 Start Root 6 End Root i

3 Start Root 2 End Root plus 3 Start Root 6 End Root i, negative 3 Start Root 2 End Root minus 3 Start Root 6 End Root i
Image with alt text: 3 Start Root 2 End Root plus 3 Start Root 6 End Root i , Image with alt text: negative 3 Start Root 2 End Root minus 3 Start Root 6 End Root i

Answers

GPT-4o mini
To find the square roots of the expression \(-36 + 36\sqrt{3}i\), we can denote the square root as \( z = a + bi \), where \( a \) and \( b \) are real numbers. We want to find \( z \) such that \( z^2 = -36 + 36\sqrt{3}i \).

Starting with the equation:

\[
(a + bi)^2 = -36 + 36\sqrt{3}i
\]

Expanding the left side gives:

\[
a^2 + 2abi - b^2 = -36 + 36\sqrt{3}i
\]

Separating the real and imaginary parts, we have:

1. Real part: \( a^2 - b^2 = -36 \)
2. Imaginary part: \( 2ab = 36\sqrt{3} \)

Now we can solve the system of equations. From the second equation, we find:

\[
ab = 18\sqrt{3} \tag{1}
\]

From the first equation, \( a^2 - b^2 = -36 \) can be rewritten as:

\[
a^2 = b^2 - 36 \tag{2}
\]

Substituting equation (1) into (2), we can express \( b \) in terms of \( a \):

\[
b = \frac{18\sqrt{3}}{a}
\]

Substituting this into \( a^2 = b^2 - 36 \):

\[
a^2 = \left(\frac{18\sqrt{3}}{a}\right)^2 - 36
\]

Simplifying this:

\[
a^2 = \frac{972}{a^2} - 36
\]

Multiplying both sides by \( a^2 \) to eliminate the fraction:

\[
a^4 + 36a^2 - 972 = 0
\]

Let \( x = a^2 \). Then the equation becomes:

\[
x^2 + 36x - 972 = 0
\]

Using the quadratic formula:

\[
x = \frac{-36 \pm \sqrt{36^2 + 4 \cdot 972}}{2}
\]

Calculating the discriminant:

\[
36^2 = 1296, \quad 4 \cdot 972 = 3888 \quad \Rightarrow \quad 1296 + 3888 = 5184
\]
\[
\sqrt{5184} = 72
\]

Plugging this back into the formula:

\[
x = \frac{-36 \pm 72}{2}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{36}{2} = 18 \)
2. \( x = \frac{-108}{2} = -54 \) (not possible since we need \( x \) to be non-negative)

So, \( a^2 = 18 \) implies \( a = 3\sqrt{2} \) or \( a = -3\sqrt{2} \).

Next, finding \( b \):

Using \( ab = 18\sqrt{3} \):

1. If \( a = 3\sqrt{2} \):
\[
3\sqrt{2}b = 18\sqrt{3} \quad \Rightarrow \quad b = \frac{18\sqrt{3}}{3\sqrt{2}} = 3\sqrt{6}
\]

2. If \( a = -3\sqrt{2} \):
\[
-3\sqrt{2}b = 18\sqrt{3} \quad \Rightarrow \quad b = -3\sqrt{6}
\]

Thus, the two square roots are:

\[
3\sqrt{2} + 3\sqrt{6}i \quad \text{and} \quad -3\sqrt{2} - 3\sqrt{6}i
\]

Matching with the provided options, the correct responses are:

\[
3\sqrt{2} + 3\sqrt{6}i, \quad -3\sqrt{2} - 3\sqrt{6}i
\]

These correspond to:

- **3√2 + 3√6 i, -3√2 - 3√6 i.**