Lots of points 53 points is the √-4 considered a real number why or why not

Mathematics

2 answers:

evablogger [386]3 years ago

4 0

No. Negative Sqaure Roots are not real numbers

Illusion [34]3 years ago

4 0

Very interesting question!

But, √-4 is not a real number; in fact, the result of √-4 is 2i.

The i is an imaginary number and it basically is a solution for √-1

(So, i^2 is = -1).

Since -4 is negative, the square root doesn't exist (in terms of real numbers) just because you can't multiply any two identical numbers to get a negative result.

-2 x -2 = +4
2 x 2 = +4

That is why √-4 is NOT considered a real number.

Hope I could help you out!
Have a good one, and God bless!

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-4x - 6y = 6 4x + 6y = -4

Sloan [31]

Step-by-step explanation:

$-4x - 6y = 6$

$4x + 6y = -4$

To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the $x$ variable when we add the two equations together.

Right now, there's a $-4x$ term in the first equation, and a $4x$ term in the second equation, so if we add those together, we'll be able to eliminate the $x$ variable altogether and solve for $y$ .

However, when we also have a $-6y$ term in the first equation and $6y$ term in the second equation, so adding these together will also eliminate the $y$ term, leaving a $0$ on the left-hand side of the equation.

If we add the two numbers on the right side of the equation, we get $-2$ , which does not equal $0$ , meaning there are no solutions to this system of equations.