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Math  /  Algebra

Question5. Find the solutions to the following equation (x+3)2=−36(x+3)^{2}=-36

Studdy Solution

STEP 1

1. The equation (x+3)2=−36(x+3)^2 = -36 involves a quadratic expression set equal to a negative number.
2. Solving the equation will involve recognizing that the square of a real number cannot be negative.
3. We will need to consider complex numbers to find solutions.

STEP 2

1. Analyze the equation to determine the nature of the solutions.
2. Solve the equation using complex numbers.

STEP 3

Analyze the equation. Recognize that the square of any real number is non-negative, and thus (x+3)2=−36(x+3)^2 = -36 has no real solutions. This indicates that the solutions must be complex.

STEP 4

To solve for xx, take the square root of both sides of the equation. Remember that when taking the square root of a negative number, we use the imaginary unit ii, where i2=−1i^2 = -1.
(x+3)2=−36(x+3)^2 = -36
Taking the square root of both sides gives:
x+3=±−36x+3 = \pm \sqrt{-36}

STEP 5

Simplify the square root of −36-36. Since −36=36⋅−1=6i\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i, we have:
x+3=±6ix+3 = \pm 6i

STEP 6

Solve for xx by isolating it on one side of the equation. Subtract 3 from both sides:
x=−3±6ix = -3 \pm 6i
This gives us the two complex solutions:
x=−3+6iandx=−3−6ix = -3 + 6i \quad \text{and} \quad x = -3 - 6i
The solutions to the equation (x+3)2=−36(x+3)^2 = -36 are:
x=−3+6iandx=−3−6i x = -3 + 6i \quad \text{and} \quad x = -3 - 6i

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