If you went to six flags with your algebra 1 class and ms. Byrd said to bring at least $25

Use complete sentences to describe why √-1 ≠ -√1

tekilochka [14]

Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

$\displaystyle \large{ {y}^{2} = x} \\ \displaystyle \large{ y = \pm \sqrt{x} }$

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

$\displaystyle \large{ {a}^{2} = a \times a = |b| }$

Where b is the result from a×a. Let's see an example.

Examples

$\displaystyle \large{ {2}^{2} = 2 \times 2 = 4} \\ \displaystyle \large{ {( - 2)}^{2} = ( - 2) \times ( - 2) = | - 4| = 4}$

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

$\displaystyle \large{ {y}^{2} = - 1}$

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

$\displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{ - 1} }$

Then where does plus-minus come from? It comes from one of Absolute Value propety.

Absolute Value Property I

$\displaystyle \large{ \sqrt{ {x}^{2} } = |x| }$

Solving absolute value always gives the plus-minus. Therefore...

$\displaystyle \large{ y = \pm \sqrt{ - 1} }$

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

$\displaystyle \large{ {y}^{2} = 1} \\ \displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{1} } \\ \displaystyle \large{ y = \pm \sqrt{1} }$

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

5 0

3 years ago

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Which is the graph of the system of equations? x + 2y = 2 4x - y = 4

Sladkaya [172]

Solve for x for x: x+2y=2
x+2y +-2y=2+-2y (add -2y on both sides.)
x=-2y+2
Substitute : -2y+2 for x in
4x-y=4:
4(-2y+2)-y=4
-9y+8=4 Simplify both sides of the equation.
-9y+8+-8=4+-8 Add -8 to both sides.
-9y=-4
-9/-9=-4-9 Divide both sides by -9
y=4/9
Substitute 4/9 for y in
x=-2y+2
x=-2(4/9)+2
x=10/9
y=4/9,x=10/9

4 0

4 years ago

Please help i need to finish this! thank you

jolli1 [7]

Answer:

x = 0

Step-by-step explanation:

0 = x²

√(0) = x

x = 0

Screenshot of Desmos graph attached showing y = x² and y = 0 intersecting.

3 0

3 years ago

Find the equaton of the line passing through the points (2,11) and (-8, -19) y= ?x ?

antoniya [11.8K]

Find the slope:

(-19 - 11) / (-8 - 2) = (-30/-10) = 3

Find the y intercept:

y = mx + b

11 = 3(2) + b

b= 5

So equation is:

y = 3x + 5

5 0

2 years ago

R+q-r; use q=6,and r=2

Anastasy [175]

Answer:

so you would replace all r with 2 which is 2+q-2

then replace the q with 6 so it would be: r+6-r

combined both of these to get

2+6-2

which would equal to 6

add the 2 and the 6 to get 8

then subtract 2 to get 6

3 0

3 years ago

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