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Math  /  Algebra

Question10 points Solve for x:(x+3)2+5=1x:(x+3)^{2}+5=1 x=−5x=-5 or -1 x=13x=13 x=−3±2ix=-3 \pm 2 i x=−3±4ix=-3 \pm 4 i

Studdy Solution

STEP 1

1. The equation to solve is (x+3)2+5=1(x+3)^{2} + 5 = 1.
2. The equation involves real and potential complex solutions.
3. Standard algebraic techniques, such as isolating the variable and solving quadratic equations, will be used.

STEP 2

1. Isolate the quadratic term.
2. Solve the resulting quadratic equation.
3. Identify the solutions and determine if they are real or complex.

STEP 3

Isolate the quadratic term (x+3)2(x+3)^{2} by subtracting 5 from both sides of the equation.
(x+3)2+5=1(x+3)^{2} + 5 = 1
Subtracting 5 from both sides:
(x+3)2=1−5(x+3)^{2} = 1 - 5

STEP 4

Simplify the right side of the equation.
(x+3)2=−4(x+3)^{2} = -4

STEP 5

Recognize that (x+3)2=−4(x+3)^{2} = -4 suggests the use of complex numbers, as the square of a real number cannot be negative.
To solve this, take the square root of both sides:
x+3=±−4x + 3 = \pm \sqrt{-4}

STEP 6

Simplify the square root of the negative number using the imaginary unit ii, where −1=i\sqrt{-1} = i.
x+3=±2ix + 3 = \pm 2i

STEP 7

Isolate xx by subtracting 3 from both sides of each equation:
x=−3±2ix = -3 \pm 2i

STEP 8

Identify the solutions as complex numbers, x=−3+2ix = -3 + 2i and x=−3−2ix = -3 - 2i.
Thus, the solutions to the equation (x+3)2+5=1(x+3)^2 + 5 = 1 are:
x=−3±2ix = -3 \pm 2i

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