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 I skipped all my classes, now I'm not being able to solve some exercises, so if you guys don't got anything to do please help solve this shit (posting procedure ofc) I'm sure this shit is really easy and you guys can solve it in minutes. And I'll tell you guys tomorrow if I managed to pass the exam.
I'll probably be here all night posting dumb questions so feel free to join and call me a faggot/retard It could be Veeky Forums's good deed /challenge but anyway here's the first one: can someone explain me how to solve this and whether it has solution?
Anthony Rogers
what is that
Sebastian Jackson
0=0
Carson Bennett
here's another pic
Austin King
>I'll probably be here all night posting
You have a pretentious attitude and I will not help you. I hope no one helps you. You are irresponsible and entitled.
Tyler Diaz
square both sides and you can solve it like normal
Parker Powell
yeah but how? I thought it could be fun for you guys too and I would be really thankful but if you don't want to don't help, but why pretentious? I just said I'm retard fag
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
Ian Butler
oops sorry its 1/a and a, i forgot to apply those exponents to the scalars 8 and 4 :)
Nicholas Anderson
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
:)
Jace Hall
put sqrt(-3) in your calculator, it won't work. sqrt(-3) is a complex number, it equals 3i.
Lincoln Scott
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
Xavier Robinson
yeah but how do you go from 8/a*4a to 1?
Austin Cruz
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
Dominic Thompson
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
Wyatt Powell
he is fucking with you, b^2 - 4ac isn't 313.
Josiah Ortiz
Did you mean [math](5 + \sqrt318)/8[/math] and [math](5 - \sqrt318)/8[/math]?
Alexander Robinson
I missed the sign and it was actually negative but the 313 part is right
Nolan Taylor
Yeah, typo on my part.
Jonathan Flores
got any other problems?
Ryan Hill
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.
Jaxson Miller
Touche
Ethan Reyes
Wtf, have any of you autists tried to solve it by hand?
Logan Jenkins
hey guys how do I solve this right, I feel like it's taking way more steps than it should
Samuel King
Yeah, I kept getting y=4 but -1!=-3.
Luke Diaz
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
Julian Phillips
multiply by the denominators, combine like terms, and solve normally
Adam Sullivan
could you spoonfeed it to me in like doing the procedure?
Matthew Long
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]
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?
Adam Ramirez
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
Asher Price
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
Leo Murphy
can anyone help me find the range of this inequality?
Landon Perez
Reported for being under 13
Jason Wilson
if I'm under 13 it should be really easy for you to explain me this
Zachary King
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)
Colton Wilson
Consequent doesn't follow from the antecedent.
Andrew Gray
Easy by guess and check.
Owen Reyes
what is there not to get about [math]\sqrt{y} - 3 - \sqrt{y} = -3 [/math]