I forgot how to do algebra (cliché, I know)

[Gascoigne]

I forgot how to do algebra (cliché, I know)

I've gone through pre-calculus, but I have somehow managed to completely forget how to handle algebra problems. My new medication makes simple math incredibly difficult (general confusion). Can you help me out with some review problems?

1) Which of the following statements are true for all x>0?

I. -5^2=25
II. 1/x + 3 = 1/(x+3)
III. cube root of x^2 = (cube root of x)^2

Is the answer A, I. only, or E, none of these? I thought it was E, but others have said A. Since there's no variable in the first equation, how can it be true for x? By the way, how do you type cube root?

2) Solve the following equation for x:
0.7x^2 + 3.2x + 1.5 = 0

I don't have any idea what I should do.

4) Which of the following statements are true for all x (does not equal) 0?

I. 1/(x+a) = 1/a + 1/x
II. 1/x + 2 = (1+2x)/x
III. 1/3 + x = 1/(3+x)

I only know how to disprove these (by plugging in numbers, basically). How can they be proven?

6) Solve the following inequality for x and express the solution set in terms of intervals:
x^2 - 2x - 5 >3

A. (-?,?)
B. (-?, 2)
C. (-2 , 4)U(4, ?)
D. (-? , -2)U(4 ,?)

Again, I can find the answer by plugging in the choices, but that's it.


There are lots more I don't understand.

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

I have somehow managed to completely forget how to handle algebra problems ... There are lots more I don't understand.

Hi Gascoigne!

Clearly, you need to take an algebra class. OR, you need to hunker down with RESOURCES and re-teach yourself about these types of problems.

We can help you get past SPECIFIC roadblocks.

We do not provide course lessons at this web site; there are "Googles" of lessons available on the Internet, and there are googols of books in libraries. If you need a teacher to guide you on how to learn all of these topics, then I really think that it's important for you to enroll in a class.

In your situation, there is NO QUICK FIX.

As far as medication issues go, I feel your pain. I suffer from schizophrenia, and I experience many situations where medication (or the lack of it) interferes with my reasoning. This means that you and I both need to WORK HARDER. I can tell when I've worked enough to get to a point of understanding where I feel comfortable in that knowledge; only you can tell whether or not you have WORKED ENOUGH to get to where you want to be.

Please, speak with your instructors and academic advisors. Make sure that you inform them of any disabilities. There are lots of "freebies" (i.e., services and accommodations) for people with medical issues, but WE need to seek them out ourselves.

Please feel free to post any specific question about a particular problem. Most of the helpers at this site are eager to provide GUIDANCE.

Cheers,

~ Mark

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

I've been refreshing with an algebra program. It's starting to come back. Can you answer the first question, though, still?

 

[Denis]

Re: I forgot how to do algebra (cliché, I know)

2) Solve the following equation for x:
0.7x^2 + 3.2x + 1.5 = 0
I don't have any idea what I should do.

This one is quite basic. If you have "no idea", then I agree 100% with Mark: only a face-to-face tutor/teacher can help.

 

[Denis]

Re: I forgot how to do algebra (cliché, I know)

1) Which of the following statements are true for all x>0?
I. -5^2=25
II. 1/x + 3 = 1/(x+3)
III. cube root of x^2 = (cube root of x)^2
Is the answer A, I. only, or E, none of these? I thought it was E, but others have said A. Since there's no variable in the first equation, how can it be true for x? By the way, how do you type cube root?

How can we help which such confusion? WHERE is "E"? Or "A"?

Cube root of x is typed: x^(1/3)

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

2) Solve the following equation for x:
0.7x^2 + 3.2x + 1.5 = 0
I don't have any idea what I should do.

This one is quite basic. If you have "no idea", then I agree 100% with Mark: only a face-to-face tutor/teacher can help.

Quadratic equation. Solved.

I've done most of the rest, too. I was just totally blanking on everything for an uncomfortable amount of time. I just did 6 from the first post, but the answers are sort of funky. Solve the following inequality for x and express the solution set in terms of intervals:
x^2-2x-5>3

becomes x^2-2x-8=0, (x-4)(x+2)=0

Now the answers, as typed on the page:
A. (-?,?)
B. (-?, 2)
C. (-2 , 4) U 4, ?)
D. (-? , -2) U (4 ,?)

Is there a reason why the sets are written like that? Rather than putting -2, 4, infinity, and negative infinity in the same set? Why a union? I know that D has to be the answer, but I don't understand the syntax.

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

Can you answer the first question, though, still?

Multiply both sides of the equation by 10

7x^2 + 32x + 15 = 0

Recognize that:

a = 7

b = 32

c = 15

Plus these values into the quadradic formula and simplify.

The two solutions for x are:

\(\displaystyle \frac{-16}{7}\;\pm\;\frac{\sqrt{151}}{7}\)

My Edit: Oops, I got mixed up on which question is "first". Oh well, did you get the same answer for #2 as I did?

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

1) Which of the following statements are true for all x>0?
I. -5^2=25
II. 1/x + 3 = 1/(x+3)
III. cube root of x^2 = (cube root of x)^2
Is the answer A, I. only, or E, none of these? I thought it was E, but others have said A. Since there's no variable in the first equation, how can it be true for x? By the way, how do you type cube root?

How can we help which such confusion? WHERE is "E"? Or "A"?

Cube root of x is typed: x^(1/3)

A, choice number one, or E, none of these choices. Commas, commas. I made a mistake with that one. The answer is choice III only.

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

Can you answer the first question, though, still?

Multiply both sides of the equation by 10

7x^2 + 32x + 15 = 0

Recognize that:

a = 7

b = 32

c = 15

Plus these values into the quadradic formula and simplify.

The two solutions for x are:

\(\displaystyle \frac{16}{7}\;\pm\;\frac{\sqrt{151}}{7}\)

That wasn't the first question - I was talking about the second one Denis responded to. I figured it out, though.

Sorry for the confusion, I've been really foggy-brained.


For clarification - I've figured out most of the problems on the sheet since posting. Sorry for the trouble. However, there is one that is still getting me:

Find the volume V of an object with is cylindrical and which has two hemispherical caps at either end. The hemispheres have a diameter of d=2/5 cm and the object has an overall length of l = 3/2 cm. (the figure on the page is shaped like a pill)

A. 53?/750 cm^3
B. 109?/750 cm^3
C. 1/2 cm^2
D. 6?/125 cm^3
E. 41?/750 cm^3

I decided to figure out the volume of the hemispheres first. I did (4/3)(?)(r^2), using 1/5 as r. I got 4?/75 as my answer. Then I did the cylinder volume, using 3/2-2/5=11/10 as the height. (?)(1/5^2)(11/10), which simplified to 11?/250. Added up, though, I got 73?/750 cm^3, which isn't an answer.

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

Which of the following statements are true for all x>0?
I. -5^2=25
II. 1/x + 3 = 1/(x+3)
III. cube root of x^2 = (cube root of x)^2

The opposite of 5^2 is NEVER equal to 5^2, and this fact has nothing to do with any value of "x".

(You are right in noticing that this equation does not even contain "x"; THEREFORE, this equation has no business being a part of this question. Whoever wrote this question is SLOPPY.)

The answer is choice III only.

Are you telling us that you believe II is NOT ALWAYS TRUE and that III is ALWAYS TRUE (for positive values of x)?

If so, then you are correct.

~ Mark

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

Which of the following statements are true for all x>0?
I. -5^2=25
II. 1/x + 3 = 1/(x+3)
III. cube root of x^2 = (cube root of x)^2

The opposite of 5^2 is NEVER equal to 5^2, and this fact has nothing to do with any value of "x".

(You are right in noticing that this equation does not even contain "x"; THEREFORE, this equation has no business being a part of this question. Whoever wrote this question is SLOPPY.)

The answer is choice III only.

Are you telling us that you believe II is NOT ALWAYS TRUE and that III is ALWAYS TRUE (for positive values of x)?

If so, then you are correct.

~ Mark

I was thinking -5 * -5. As far as I know that's what -5^2 means. My teachers must be lazy, because I've always seen -5^2 as interchangeable with (-5)^2.

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

Ah ha!

-5^2 does NOT mean -5 * -5.

-5^2 means -1 * 5^2

The way to indicate -5 * -5 is (-5)^2

-5^2 is NOT interchangable with (-5)^2

 

[Gascoigne]

Re: I forgot how to do algebra (cliché, I know)

Ah ha!

-5^2 does NOT mean -5 * -5.

-5^2 means -1 * 5^2

The way to indicate -5 * -5 is (-5)^2

-5^2 is NOT interchangable with (-5)^2

Whoops.

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

Is the "whoops" because you suddenly realized that your teachers are not so lazy afterall? (That's a rhetorical question.) :wink:

... I decided to figure out the volume of the hemispheres first. I did (4/3)(?)(r^2), using 1/5 as r. I got 4?/75 as my answer.

Volume is always three-dimensional (cubed); area is always two-dimensional (squared).

Try r^3 instead because r^2 makes the volume formula incorrect!

Cheers,

~ Mark

 

[mmm4444bot]

Re: I forgot how to do algebra (cliché, I know)

... Now the answers, as typed on the page:
A. (-?,?)
B. (-?, 2)
C. (-2 , 4) U 4, ?)
D. (-? , -2) U (4 ,?)

Is there a reason why the sets are written like that? Rather than putting -2, 4, infinity, and negative infinity in the same set? Why a union? I know that D has to be the answer, but I don't understand the syntax.

You're pretty clever by figuring out how to post an infinity symbol.

Firstly, let's clarify that the values of x that make the expression greater than 5 are not the following set.

{-?, -2, 4, ?}

Infinity is not even a number, so it can't be a value for x.

The notation in this multiple-choice question is called Interval Notation. When there are infinite solutions to an inequality (i.e., infinite values on the number line which make the inequality statement true when substituted for x), we have to write the solution as an interval or an aggregate of intervals.

Regarding your original inequality, the values on the number line from -2 through 4 are NOT solutions because those values do not satisfy the inequality. There is a gap in the solution "line". This is why we need to use a union of two sets.

(-? , -2) U (4 ,?)

This solution in Interval Notation is read as, "The set of all x such that x is a number less than -2 or a number greater than 4".

Check out "interval notation" at Google; make sure you understand when to begin and end intervals with ( and ) and [ and ].

Cheers,

~ Mark