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Ksenya-84 [330]
3 years ago
13

What is the solution of the equation when solved over the complex numbers? x^2+27=0

Mathematics
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

(0,27)

Step-by-step explanation:

x2 + 27 = 0

Solutions based on quadratic formula:

x1  

= −0 − √ 02 − 4×1×27   2×1 = 0 − 6 × √ 3 i   2 ≈ −5.196152i

x2  

= −0 + √ 02 − 4×1×27   2×1 = 0 + 6 × √ 3 i  2 ≈ 5.196152i

Extrema:

Min = (0, 27)

juin [17]3 years ago
7 0

Answer:

x =i\sqrt{27}

Step-by-step explanation:

x^2+27=0

To solve for x, we need to get x alone

x^2+27=0

Subtract 27 from both sides

x^2+27-27=0-27

x^2=-27

To remove square root , take square root on both sides

\sqrt{x^2} =\sqrt{-27}

x =\sqrt{-27}

The value of square root (-1) is 'i'

x =i\sqrt{27}

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Please help!
Gemiola [76]

The derivative of f(x) = 2\cdot x^{2}-9 is f'(x) = 4\cdot x.

In this exercise we must apply the definition of derivative, which is described below:

f'(x) =  \lim_{x \to 0} a_n \frac{f(x+h)-f(x)}{h} (1)

If we know that f(x) = 2\cdot x^{2}-9, then the derivative of the expression is:

f'(x) =  \lim_{h \to 0} \frac{2\cdot (x+h)^{2}-9-2\cdot x^{2}+9}{h}

f'(x) = 2\cdot \lim_{h \to 0} \frac{x^{2}+2\cdot h\cdot x + h^{2}-2\cdot x^{2}}{h}

f'(x) = 2\cdot  \lim_{h \to 0} 2\cdot x + h

f'(x) = 4\cdot x

The derivative of f(x) = 2\cdot x^{2}-9 is f'(x) = 4\cdot x.

We kindly invite to check this question on derivatives: brainly.com/question/23847661

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