Asked by yeslian c
Which is the solution to the equation 2(x−16)2=−8 , re-expressed as a complex number in the standard form a±bi ?(1 point)
Responses
16±2i
16 plus or minus 2 i
16±2i2–√
16 plus or minus 2 i Start Root 2 End Root
4±2i2–√
4 plus or minus 2 i Start Root 2 End Root
−4±2i
Responses
16±2i
16 plus or minus 2 i
16±2i2–√
16 plus or minus 2 i Start Root 2 End Root
4±2i2–√
4 plus or minus 2 i Start Root 2 End Root
−4±2i
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the equation \( 2(x - 16)^2 = -8 \), we can start by isolating the term involving \( x \):
1. Divide both sides by 2:
\[
(x - 16)^2 = -4
\]
2. Taking the square root of both sides gives:
\[
x - 16 = \pm \sqrt{-4}
\]
3. Since the square root of \(-4\) can be expressed in terms of imaginary numbers (using \( i \)), we have:
\[
\sqrt{-4} = \sqrt{4} \cdot i = 2i
\]
4. Therefore, we rewrite the equation as:
\[
x - 16 = \pm 2i
\]
5. Now, solve for \( x \):
\[
x = 16 \pm 2i
\]
Thus, the solutions to the equation in the standard form \( a \pm bi \) are:
\[
16 \pm 2i
\]
So, the correct response is:
**16 ± 2i** (or **16 plus or minus 2 i**).
1. Divide both sides by 2:
\[
(x - 16)^2 = -4
\]
2. Taking the square root of both sides gives:
\[
x - 16 = \pm \sqrt{-4}
\]
3. Since the square root of \(-4\) can be expressed in terms of imaginary numbers (using \( i \)), we have:
\[
\sqrt{-4} = \sqrt{4} \cdot i = 2i
\]
4. Therefore, we rewrite the equation as:
\[
x - 16 = \pm 2i
\]
5. Now, solve for \( x \):
\[
x = 16 \pm 2i
\]
Thus, the solutions to the equation in the standard form \( a \pm bi \) are:
\[
16 \pm 2i
\]
So, the correct response is:
**16 ± 2i** (or **16 plus or minus 2 i**).
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