Subtract And Simplify 85i 5 8i2 Simplify I63

Subtract And Simplify 85i 5 8i2 Simplify I63

Subtract And Simplify 85i 5 8i2 Simplify I63

Given the set of algebraic problems involving complex numbers, quadratic equations, function analysis, and algebraic manipulation, this paper provides detailed solutions and explanations to all tasks for comprehensive understanding.

Paper For Above instruction

1. Subtract and simplify: (−8 + 5i) − (−5 − 8i)

To perform the subtraction of the complex numbers, we distribute the minus sign across the second parentheses:

(−8 + 5i) − (−5 − 8i) = (−8 + 5i) + (5 + 8i)

Next, combine like terms:

(−8 + 5) + (5i + 8i) = (−3) + (13i)

Therefore, the simplified form is:

−3 + 13i

2. Simplify: (−i)^6

The powers of complex numbers involving i follow periodicity, with i^4 = 1. To simplify (−i)^6, note that:

(−i)^6 = [ (−1)  i ]^6 = (−1)^6  i^6 = 1 * i^6

Recall that:

i^6 = i^{4 + 2} = i^4  i^2 = 1  (−1) = -1

Thus,

(−i)^6 = -1

3. Determine the nature of the solutions of the equation: x^2 - x + 8 = 0

The quadratic equation x^2 - x + 8 = 0 can be analyzed using the discriminant:

D = b^2 - 4ac = (−1)^2 - 4  1  8 = 1 - 32 = -31

Since D

Explicitly, using quadratic formula:

x = [−b ± √D] / 2a = [1 ± √(−31)] / 2 = [1 ± i * √31] / 2

Thus, the solutions are: x = (1/2) ± (i * √31) / 2

4. Multiply and simplify: [x - (5+9i)] * [x - (5−9i)]

The expression is a product of conjugates, which simplifies using the difference of squares:

[x - (5+9i)]  [x - (5−9i)] = [x - 5 - 9i]  [x - 5 + 9i] = [(x - 5)^2 - (9i)^2]

Since (9i)^2 = 81 i^2 = 81 (−1) = -81, this becomes:

(x - 5)^2 - (−81) = (x - 5)^2 + 81

Expanding (x - 5)^2:

x^2 - 10x + 25 + 81 = x^2 - 10x + 106

The simplified form is:

x^2 - 10x + 106

5. Use the quadratic formula to find the exact solutions of: x^2 + 6x + 4 = 0

Discriminant computation:

D = 6^2 - 4  1  4 = 36 - 16 = 20

Solutions:

x = [−6 ± √20] / 2 = [−6 ± 2√5] / 2 = [−3 ± √5]

Thus, the solutions are x = −3 + √5 and x = −3 − √5.

6. Solve: x - 9√x + 8 = 0

Let √x = t, which implies x = t^2. Substitute into the equation:

t^2 - 9t + 8 = 0

This quadratic in t can be solved using the quadratic formula:

t = [9 ± √(81 − 32)] / 2 = [9 ± √49] / 2 = [9 ± 7] / 2

Solutions for t:

t = (9 + 7)/2 = 16/2 = 8;   t = (9 - 7)/2 = 2/2 = 1

Recall that t = √x, so:

√x = 8 ⇒ x = 64


√x = 1 ⇒ x = 1

Both are valid as square root functions are non-negative in real numbers. Final solutions:

x = 64 and x = 1

7. For f(x) = -8x^2 - 40x + 3, find the vertex, determine whether there is a maximum and its value, the range, and increasing/decreasing intervals

Vertex: The vertex of a quadratic f(x) = ax^2 + bx + c is at x = -b/2a:

x_v = -(-40) / (2 * -8) = 40 / -16 = -2.5

Calculate f(-2.5):

f(-2.5) = -8(-2.5)^2 - 40(-2.5) + 3


= -8(6.25) + 100 + 3


= -50 + 100 + 3 = 53

The vertex is at (-2.5, 53), which is a maximum because the parabola opens downward (leading coefficient negative).

Maximum value of f(x): 53.

The parabola's range:

(-∞, 53]

Intervals of increase and decrease:

  • Decreasing on (−∞, -2.5)
  • Increasing on (−2.5, ∞)

8. Solve: √x + 49 = x - 7

Isolate √x:

√x = x - 7 - 49 = x - 56

Square both sides to eliminate the root:

x = (x - 56)^2 = x^2 - 112x + 3136

Bring all terms to one side:

x^2 - 113x + 3136 = 0

Quadratic in standard form. Discriminant:

D = (−113)^2 - 4  1  3136 = 12769 - 12544 = 225

Solutions:

x = [113 ± √225] / 2 = [113 ± 15] / 2

Solutions:

x = (113 + 15)/2 = 128/2 = 64


x = (113 - 15)/2 = 98/2 = 49

Check for extraneous solutions by substituting back:

  • For x=64:
√64 + 49 = 8 + 49 = 57, and x -7 = 64 -7 = 57 ⇒ valid.
  • For x=49:
    • √49 + 49 = 7 + 49 = 56, and x -7 = 49 -7 = 42 ≠ 56 ⇒ extraneous.

    Answer: x=64

    9. Solve (1/A) = (1/s) + (1/t) for t

    Express t in terms of A and s:

    1/t = 1/A - 1/s
    

    1/t = (s - A) / (A * s)

    Take the reciprocal to solve for t:

    t = (A * s) / (s - A)

    Final solution:

    t = (A * s) / (s - A)

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