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3 years ago

The solution to an equation is

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

5 0

Answer:

solution is an assignment of values to the unknown variables that makes the equality in the equation true. In other words, a solution is a value or a collection of values (one for each unknown) such that, when substituted for the unknowns, the equation becomes an equality.

Step-by-step explanation:

I'm not really sure wym but this might help

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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

2 years ago

Read 2 more answers

A company developed the following linear model to help budget annual expenses over time. In this model, x represents the number

ollegr [7]

The only reasonable answer I could think for this is "A.The company's expense budget for 2009 was $189,785."
I believe this because since x is the years after 2009 then you wouldn't times that by the expense budget for that year or any year eliminating B and C. When finding the total expense budget for this you would have to take in account the budget from 2009 so that would lead it to add 189,758. Since we are talking about after 2009 then D would be eliminated since 2008 is before 2009.
I hope this helps! Good luck!

6 0

2 years ago

Please help work this equation

Flauer [41]

Hi!

Given these two equations:

$x = y + 8$

$x + 4 = 2(y + 4)$

We want to solve using the substitution method. Knowing that x is equal to y + 8, we can simply plug in 'y + 8' in for x in the second equation, like so:

$(y+8)+4=2(y+4)$