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3 years ago

What is the solution of the equation when solved over the complex numbers? X2+24=0.

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

8 0

Answer:

In the picture

zlopas [31]3 years ago

5 0

Answer:

$x=+-2i\sqrt{6}$

Step-by-step explanation:

$x^2+24=0$

To solve for x we need to get x alone

Subtract 24 from both sides

$x^2+24=0$

$x^2=-24$

Now take square root on both sides to remove the square

$x^2=-24$

$\sqrt{x^2} =\sqrt{-24}$

$x=+-2\sqrt{-6}$

square root of -1 is 'i'

$x=+-2i\sqrt{6}$

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3 years ago

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2 years ago

Find a power series for the function, centered at c. g(x) = 4x x2 2x − 3 , c = 0

BartSMP [9]

The power series for given function $g(x)=\frac{4x}{(x-1)(x+3)}$ is $g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$

For given question,

We have been given a function g(x) = 4x / (x² + 2x - 3)

We need to find a power series for the function, centered at c, for c = 0.

First we factorize the denominator of function g(x), we have:

$\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}$

We can write g(x) as,

$\Rightarrow g(x)=\frac{1}{x-1}+\frac{3}{x+3}\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1+\frac{x}{3} }\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1-(-\frac{x}{3} )}\\$

We know that, $\frac{1}{1-x}=\sum{_{n=0}^\infty}~{x^n}$ if |x| < 1

and $\frac{1}{1-(-\frac{x}{3} )}=\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n$ if $|\frac{x}{6}| < 1$

$\Rightarrow g(x)=-\sum{_{n=0}^\infty}~x^n+\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n\\$ if |x| < 1 and if $|\frac{x}{6}| < 1$

$\Rightarrow g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$ if |x| < 1

Therefore, the power series for given function $g(x)=\frac{4x}{(x-1)(x+3)}$ is $g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$

Learn more about the power series here:

brainly.com/question/11606956

#SPJ4

5 0

1 year ago