HURRY. Write a ratio for the following word phrase. Write the fraction in lowest terms.

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

Read 2 more answers

2) 4 6 2 3 If L= | 5 8 and M= 1 4 3 -2 -5 3 3 Find L-M -)) 2 4 8 A ) 3 4 5 2 3 B) 4 4 8 -5 6 9 o 6 12 -2 1 6 6 2 D) 9 12 1

Misha Larkins [42]

$\\ \sf\longmapsto L-M$

Put values

$\sf \left[\begin{array}{cc}\sf 4&\sf 6\\ \sf 5 &\sf 8 \\ \sf 3 &\sf -2\end{array}\right]-\left[\begin{array}{cc}\sf 2&\sf 3\\ \sf 1 &\sf 4 \\ \sf -5&\sf3\end{array}\right]$

Just substract corresponding terms

$\\ \sf\longmapsto \left[\begin{array}{cc}\sf 2 &\sf 3\\ \sf 4&\sf4\\ \sf 8&\sf -5\end{array}\right]$

Option B

6 0

3 years ago

During Raul’s exercise program, he recorded how much weight he lost or gained each week. After 5 weeks, Raul had collected the d

Amiraneli [1.4K]

The correct answer for the question that is being presented above is this one: "A. The total of Raul’s numbers is not 0."During Raul’s exercise program, he recorded how much weight he lost or gained each week. After 5 weeks, Raul had collected the data shown in the chart.

4 0

3 years ago

The sum of two.number is.24. the.sun.of the square.of the two.nunber is 306. what is the product.of the two.number

tresset_1 [31]

A) x + y = 24
B) x^2 + y^2 = 306

A) x = 24 -y
Then substituting this into B)
(24 - y)^2 +y^2 = 306
576 -48y +y^2 + y^2 = 306
2 y^2 -48y + 270 = 0
x1 = 15
x2 = 9

3 0

3 years ago

Solve for a. 2z=4a/m +3

Finger [1]

Answer:

Step-by-step explanation:

2z = 4a/m +3

-3 -3

2z - 3 = 4a/m

×m ×m

(2z -3)m = 4a

÷4 ÷4

[(2z -3)m]/4 = a

a = (2mz - 3m)/4

4 0

3 years ago