[High School - Algebra II] Radical Expressions & Exponents, etc.

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

To solve any quadratic just use the quadratic formula. Any other method is useless unless she requires you to do something specific (which sucks but unless you try to explain it I can't really help).

So for ax^2+bx+c = 0, x = (-b+/-(b^2-4ac))/(2a)

For the others what you're trying to do is find squares within the larger numbers and remember that the powers add, not multiply when it's like (3)^(1/2)*(3)^(1/3) = (3)^(1/2+1/3) but when it's (3^2)^(1/2) = 3^(2*1/2).

So, for like 8, 75/3 = 25 and 5*5 = 25, so there's a 3*5^2 in the first number. And 3*9 = 27 and 3*3=9 so there's a 3*3^2 in the second number. Since there's a common factor of (3)^(1/2) in both the numbers and the other squares just reduce you get (3)^(1/2)(4*5+3) = 23*(3)^(1/2).

It's really just a matter of breaking down the numbers into their lowest factors or whatever I don't remember the terms exactly but do you get what I mean?

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

Lol, blu. Nicely put.

For number 2, roots and powers tend to 'cancel' each other out (just like multiplication and devision, per se). Just like:

2 * 1/2 = 1 (You know 1/2 is also 1 devided by 2, etc etc.)

I'm not really good at wording this out but powers and roots have the same concept:

= a^5/5 * b^2/5

EDIT: For numbers 4, 16, and 17 use the quadratic formula, just like what blu said. Since the answers were given, it's clear you are to solve for 'x'. Use this:



Where:



Oh, the plus or minus sign states that x has two answers, a negative and a positive one.

EDIT2: For numbers 10 & 12, use the same concept just like the one for number 2, but the other way around. Got me? ;]

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

Ok, so you told me to do the opposite for numbers 10 and 12 and I did and I got the answer for number 12 but it's wrong.

I'm so confused now! xD

I pretty much get everything else now though, thanks.

I was writing the quadratic forumla wrong, lmao. .__.'

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

For 12, (27)^(1/3) = (3^3)^(1/3)= 3^(3/3) = 3, and you can do the other half of the simplification.

Just remember that (x^a)^b=x^(a*b) and x^a*x^b=x^(a+b).

Also, if you ever forget the quadratic formula, you can solve for the roots by completing the square, if you learned that, but I don't think I learned that in Algebra II, probably Pre-Calc.

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

OH YEAH, that totally went past my mind when I was working it out in my head. ^^' Yeah that makes more sense now, thanks so much.

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

Oh, protip, don't do math in your head. Always write it all out.

The more methodical you are, the less likely you are to make mistakes. And it's also easier to find your mistakes.

Re: [High School - Algebra II] Radical Expressions & Exponents, etc.

For your education as much as my own, I'm going to try to derive the quadratic equation right here. (Only the positive square root is indicated in the final equation, when you know as well as I do that what really belongs there is a combination plus/minus sign that I don't know how to do.)

ax^2 + bx + c = 0

ax^2 + bx = -c

x^2 + (b/a)x = -(c/a)

x^2 + (b/a)x + (b^2/4a^2) = (b^2/4a^2) - (c/a)

c/a = 4ac/4a^2, so we end up with ((b^2) - 4ac)/4a^2 on the right side after combining the terms.

(x + (b/2a))^2 = ((b^2) - 4ac)/4a^2

x + (b/2a) = sqrt(b^2 - 4ac)/2a

x = (-b + sqrt(b^2 - 4ac))/2a