So ok Veeky Forums I need your help I have an exam tomorrow and I thougt it was gonna be easy because it was algebra so...

can you change sqrt (y -3) to sqrt (y) sqrt (-3) even if the 3 isn't multiplying?

just multiply both sides by [math]sqrt{y}[/math] then everything cancels out.

multiply both sides by 0

[math]\left\sqrt{\sqrt{8}+3}-1\right^{2} = 2[/math]

I put it in wrong
[math]( \sqrt{ \sqrt{8} + 3} - 1 )^{2} = 2[/math]

nah they trolling u

just think of sqrt as ^2

you can't do: (a+b)^2 = a^2 + b^2, so you can't do sqrt(a+b) = sqrt(a) + sqrt(b) either.

if ever in doubt, just sub some numbers in for a and b to see if they equal each other, eg

a=16
b=64

sqrt(16 + 64) = sqrt(16) + sqrt (64)
sqrt(80) = 4 + 8
8.94 =/= 12

Ok guys thanks a lot I think i got it, next one is this it says to simplify and that the answer is 1

for problems like the first, its usually better to just rewrite sqrt as ^(1/2), makes it easier to deal with.

for Q2 you would do: (a^x)^y = a^(x*y)

(8a^3)^(-1/3) = 8a^(3*-1/3) = 8a^(-3/3) = 8a^(-1) = 8/a

(4a^2)^(2/4) = 4a^(2*(2/4)) = 4a^(4/4) = 4a^(1) = 4a

here are some common exponent laws:
a^m * a^n = a^m+n
a^m / a^n = a^m-n
(a^m)^n = a^m*n

oops sorry its 1/a and a, i forgot to apply those exponents to the scalars 8 and 4 :)

Oh if those are being multiplied, include a dot between them to indicate that.

Then it would be 1/a * a = 1 * (a/a) = 1 * (a^1 * a^-1) = 1 * (a^1-1) = 1 * a^0 = 1 * 1 = 1

:)

put sqrt(-3) in your calculator, it won't work. sqrt(-3) is a complex number, it equals 3i.

ok thanks guys I feel at this pace I might be able to save the class, this one seems kinda complex but is multiple choice so maybe I'm supposed to deduce the answer without solving it. The solutions are:
A) it has two solutions: 2 and -2
B) it has one solution: 2
C) it has two solutions 5+sqrt(313)/8 and 5+sqrt(313)/8
D) it has no solution

yeah but how do you go from 8/a*4a to 1?

You can only solve this if your definition of sqrt allows you to choose the positive root, or the negative root. If you can, then y = 4, but sqrt(y-3) = -1, and - sqrt(y) = -2

multiple choice are great because you can usually get rid of the obviously wrong answers. For example, it can't be 2 or -2 because it will have one denominator equaling zero.

without doing the working out, it should be C (though it should be 5+sqrt(313)/8 or 5-sqrt(313)/8). What you do is multiply the top by the bottom, gather like terms and then use the quadratic formula

he is fucking with you, b^2 - 4ac isn't 313.

Did you mean [math](5 + \sqrt318)/8[/math] and [math](5 - \sqrt318)/8[/math]?

I missed the sign and it was actually negative but the 313 part is right

Yeah, typo on my part.

got any other problems?

It is amazing how few people think to simply plug this into wolfram...

no solution kohai

you can come up with solutions, yes but when you double check them with the og question they turn out to be extraneous.

Touche

Wtf, have any of you autists tried to solve it by hand?

hey guys how do I solve this right, I feel like it's taking way more steps than it should

Yeah, I kept getting y=4 but -1!=-3.

And a kinda off topic but related question, my exam is tomorrow at 8:00 am how long should I sleep or should I just skip it altogether I do have some modafinil pills just in case

multiply by the denominators, combine like terms, and solve normally

could you spoonfeed it to me in like doing the procedure?

Please ignore, I'm trying this neat math feature.

[math]\sqrt (y-3) - \sqrt y = -3[/math]
[math]\sqrt (y-3) =\sqrt y -3[/math]
[math](\sqrt (y-3))^2 = (\sqrt y - 3)^2[/math]
[math]y - 3 = y - 6\sqrt y + 9[/math]
[math]-12 = -6\sqrt y[/math]
[math]2 = \sqrt y[/math]
[math]2^2 = (\sqrt y)^2[/math]
[math]4 = y[/math]

but

[math]\sqrt((4)-3) - \sqrt(4) \neq -3[/math]
[math]-1 \neq -3[/math]

...

depends.

sleep if you know the material and have practiced

if not, stay up and do what you can. don't try to learn all the problems. focus on a select few you know will be on the test that you think you can get, and make sure you know how to solve them.

Usually around 60% of the test will be straight forward questions which test basic concepts/knowledge, these are what you want to learn tonight.

Do you know which question types or chapters will be on the test?

what you want to do here is get x in the numerator to make it easy to solve.

to do this, you simply multiply the left and right hand sides by each x term.

so you'd multiply the whole equation by (x+3) and (3x+7) and (x+1), leading to cancellations of the denominator terms.

so you'd end up with
2*(3x+7)(x+1) = 3(x+3)(x+1) + 1(x+3)(3x+7)
>then simplify
6x^2+20x+14 = 3x^2+12x+9 + 3x^2+16x+21
>then you move every x term to the LHS and constants to the RHS, and group like terms for easy accounting
(6x^2 - 6x^2) + (20x - 12X - 16X) = 9 + 21 - 14
>then solve for x
0 - 8x = 16, so x = -2

yes I do have the topics of the test but they're really vague terms so I'm just doing past exams and asking you guys for help and hoping for the best

can anyone help me find the range of this inequality?

Reported for being under 13

if I'm under 13 it should be really easy for you to explain me this

you know what the range is, yeah?

expand the LHS, you'll find out that it's a quartic equation with roots at -4, 0, 1 and 3. Use that to find the range. (Hint: it's y < 3)

Consequent doesn't follow from the antecedent.

Easy by guess and check.

what is there not to get about
[math]\sqrt{y} - 3 - \sqrt{y} = -3 [/math]