Picking numbers: expression is odd integer ("2(2)^2 + 1")

[Illvoices]

2(2)²+1

Where do I start. With the parenthesis times the exponent or the constant times the exponent. Please help i need to do some practices for my SAT

 

[Dr.Peterson]

2(2)²+1

Where do I start. With the parenthesis times the exponent or the constant times the exponent. Please help i need to do some practices for my SAT

If you're asking about how to evaluate this expression, you have to follow the order of operations: apply the exponent first (to just the base, 2), and then multiply the result by the multiplier (the first 2), and finally do the addition.

What do you get?

And what was the entire problem? Your title is not clearly relevant.

 

[Illvoices]

Ok so when I follow the order of operations i get. 9

The question was which of the following expression s will produce an odd number for any integer a?

a) a²
b) a²+1
c)2a²+1
d)3a²+2
e)4a²​+4

 

[Dr.Peterson]

Ok so when I follow the order of operations i get. 9

The question was which of the following expressions will produce an odd number for any integer a?

a) a²
b) a²+1
c)2a²+1
d)3a²+2
e)4a²​+4

You are correct about the evaluation; so choice (c) does produce an odd number at least for a=2.

Can you convince yourself that its value will be odd for any integer a? Why?

And can you see why the others can all produce an even number, so that they can't be the answer? All you need there is to give an example for each.

 

[Steven G]

2(2)²+1

Where do I start. With the parenthesis times the exponent or the constant times the exponent. Please help i need to do some practices for my SAT

Just read 2(2)² and do NOT worry about order of operations. 2(2)² is read as 2 times 2squared or 2 times 2 to the 2nd power. The times symbol separates what you are multiply. To the left of the times symbol is 2 which of course is just 2. To the right of the times symbol is 2² which is 4. So multiply 2 and 4 and get 8 and move onto the next problem.

To learn how to do these type problems please view this video https://www.youtube.com/watch?v=fIKqQbRtZJw&index=9&list=PLtmopxjCQWixC0DvQqrROOLqKE0zLAmyc

 

[Illvoices]

U still don't understand how to solve these problems can someone teach me. Or give me another example of a link with these problems. Where I can learn how to solve them.

I still don't understand what they asking me in these problems.

 

[JeffM]

U still don't understand how to solve these problems can someone teach me. Or give me another example of a link with these problems. Where I can learn how to solve them.

I still don't understand what they asking me in these problems.

Integers are either even or odd.

An even integer can be expressed as 2 times an integer. So it is easy to demonstrate that the sum of two even numbers is even:

\(\displaystyle a \text { and } b \text { are even } \implies a = 2m \text { and } b = 2n \implies a + b = 2m + 2n = 2(m + n) \text { is EVEN.}\)

An odd number can be expressed as 2 times an integer plus 1. So it is easy to show that the product of two odd numbers is odd.

\(\displaystyle a \text { and } b \text { are odd } \implies a = 2m + 1 \text { and } b = 2n + 1 \implies\)

\(\displaystyle ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(mn + m + n) + 1 \text { is ODD.}\)

So what this problem is asking you to do is to determine whether the given expression is odd whether or not the unknown number used in the expression is odd or even. That means you need to test two cases for each expression. For example:

\(\displaystyle a \text { is odd } \implies a = 2m + 1 \implies\)

\(\displaystyle a^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1 \text { is ODD.}.\)

\(\displaystyle a \text { is even } \implies a = 2m \implies\)

\(\displaystyle a^2 = (2m)^2 = 4m^2 = 2(2m^2) \text { is EVEN.}[/text]

So a squared is not always odd.\)

 

[Steven G]

You are correct about the evaluation; so choice (c) does produce an odd number at least for a=2.

Can you convince yourself that its value will be odd for any integer a? Why?

And can you see why the others can all produce an even number, so that they can't be the answer? All you need there is to give an example for each.

1)Sum of 2 odds numbers are even
2)Sum of 2 even numbers are even
**) sum of odd and even is odd
3)If you square different integers you will get some even results and some odd results
4)Every even number can be written as 2 times an integer (so an even*even=even and odd*even=even)
5) odd*odd is odd

Using these 6 rules above we can look at each choice and either reject it or accept it.

a²: we reject by rule 3
a²+1: we do not know if a² is even or odd, so when we add 1 to a² (to get a²+1)we still don't know if it is even or odd, so we reject it.
2a²+1: Since 2a² is even by rule 4, when we add 1 we get odd. So 2a²+1 is odd.
3a²+2: By rule 3, we know that a² can be either even or odd. If a² is even and you multiply by 3 (by rule 4) you get an even and if you add 1, it becomes odd. So we reject 3a²+2
4a²​+4: 4a² is even (since even * odd or even = even). 4 is even. The sum of these two even number by rule 2 is even. THIS IS THE ANSWER

 

[stapel]

U still don't understand how to solve these problems

The helper understands these questions perfectly well. I'm sorry that you're so lost that you can't understand the helps you've been given. :shock:

can someone teach me.

Unfortunately, it is not reasonably feasible to attempt hours of classroom instruction within this environment. Also, you are posting questions related to symbolic algebraic thinking, but seem not to understand the order of operations, which is something that is covered in pre-algebra. So it appears that you're missing months, if not years, of material.

I still don't understand what they asking me in these problems.

Okay; let's start at the beginning: Do you understand what "odd" and "even" numbers are?

If so, do you understand how to evaluate variable expressions?

Thank you!