Math again? Google isn't helpful

[Sandform]

Amiright? Find the intervals on the real number lin for which the radicand is nonnegative(greater than or equal to zero). 1. Sqrt(x-5) 2. (7-2x)^(1/4) I don't understand the question...but my answers are this. 1. X>5 2. X<3.5 or X<7/2

[archdreamer]

I have no idea what a 'radican' is, but they seem right, except that those less than or equal to and more than or equal to signs need to be just less than and more than signs. The expressions equal to zero for X=5 and X=3.5 respectively, and are undefined over the reals for X less than those (but still defined over the complex numbers).

[Sandform]

I have no idea what a 'radican' is, but they seem right, except that those less than or equal to and more than or equal to signs need to be just less than and more than signs. The expressions equal to zero for X=5 and X=3.5 respectively, and are undefined over the reals for X less than those (but still defined over the complex numbers). Sorry I meant to type Radicand. A radicand is any number under the radical such as The symbol √ used to indicate the square root or nth root. I think they need to be greater/less than or equal to because the actual equation (based on the question posed before the equation) is... √(x-5)>0 Oh wait I think I get what you're saying.

[drewmandan]

I have no idea what a 'radican' is, but they seem right, except that those less than or equal to and more than or equal to signs need to be just less than and more than signs. The expressions equal to zero for X=5 and X=3.5 respectively, and are undefined over the reals for X less than those (but still defined over the complex numbers). Wrong. He requires greater than or equal to zero, so the signs require equality as well. Your comment about real versus imaginary is correct but quite irrelevant to the problem as stated.

[archdreamer]

Sandform: Did you edit your post? I could have sworn it said 'greater than zero'. Oh well, what drew says is correct, if we are trying to solve √(x-5)>0 for x.

[NeAvO]

Your answers are correct: x-5>o x>5 Pretend thats got a line under it. (7-2x)(1/4)>o 28-8x>o 28>8x 28/8>x 14/4 7/2 3.5 And hell, I don't even have to worry about maths again. Life is good

[drewmandan]

I, on the other hand, am about to begin third year university math. I'm gonna be finding so many eigenvectors that I'll be speaking German by April.

[Sandform]

Cool! Math is fun when you get back into it for a while hehe. It is just jumping in after half a year of non-usage that is annoying for the first week or so.

[drewmandan]

Cool! Math is fun when you get back into it for a while hehe. It is just jumping in after half a year of non-usage that is annoying for the first week or so. Yeah, especially when you're jumping into partial differential equations and real analysis and other goodies...

[wendylove]

I, on the other hand, am about to begin third year university math. I'm gonna be finding so many eigenvectors that I'll be speaking German by April I thought you did Physics? I hate working out eigenvalues. Yeah, especially when you're jumping into partial differential equations and real analysis and other goodies... I thought first year was real analysis? Not the third year. LOL. Sandform, does you're anwser agree with the book? I hate partial differential equations, the worst thing ever. I hate intergration, because differentiation is soo easy.

[drewmandan]

I thought you did Physics? I do do physics, but my major is math. I thought first year was real analysis? Not the third year. LOL. Uh, no. I'm talking about Lebesgue integrals and proofs of stuff. Way way above first year. I hate partial differential equations, the worst thing ever. I hate intergration, because differentiation is soo easy. Spoken like a true first year.

[Sandform]

Sandform, does you're anwser agree with the book? I wasn't issued a book sadly. Nor did my teacher give us notes hehe. She expects us to know this already, which is reasonable, considering it is basic algebra and I'm taking calculus.

[drewmandan]

Ok, here's a more challenging one: x^2 - x -6 >=0 Hint: You might want to draw the function to get your bearings, otherwise it might be difficult to come up with the full answer.

[Sandform]

Ok, here's a more challenging one: x^2 - x -6 >=0 Hint: You might want to draw the function to get your bearings, otherwise it might be difficult to come up with the full answer. You talking to me? Spoiler for answer: (1+-√(1-4*1*-6)) / 2 (1 +- √(25)) /2 (1 +- 5) / 2 -4 / 2 = -2 6 / 2 = 3 x = -2 x = 3 Lets turn this into a math game in which the next question is harder than the last...and you have to show your work. Here's a more challenging one. √(3x-2) = 4 - x

[drewmandan]

You talking to me? Spoiler for answer: (1+-√(1-4*1*-6)) / 2 (1 +- √(25)) /2 (1 +- 5) / 2 -4 / 2 = -2 6 / 2 = 3 x = -2 x = 3 This is precisely why I suggested you draw the function. Your answer is incomplete. Lets turn this into a math game in which the next question is harder than the last...and you have to show your work. Here's a more challenging one. √(3x-2) = 4 - x Yuck. √(3x-2) = 4 - x 3x -2 = 16 - 8x + x^2 x^2 - 11x +18 =0 x= (11 ± √(121-72))/2 x = 11/2 ± 7/2 x = 9, x = 2 edited for stupidy Here's my response! x^3 + 6x^2 +11x +6 >= 0

[Sandform]

This is precisely why I suggested you draw the function. Your answer is incomplete. Oops I meant Spoiler for oops.: x < -2 x > 3 I didn't see that you wrote > before =. also... √(3x-2) = 4 - x √(3*12.5-2) = 4 - 12.5 ~5.9581 = -8.5 Also -1.5(3) - 2 results in √(-6.5) So that can't be right either...

[drewmandan]

Oops I meant Spoiler for oops.: x < -2 x > 3 I didn't see that you wrote > before =. also... √(3x-2) = 4 - x √(3*12.5-2) = 4 - 12.5 ~5.9581 = -8.5 Correct. Now do the one I just posted, and I'll give you a cookie.

[drewmandan]

Crap I made a mistake. I forgot to divide the 7 by 2. Fixed it.

[Sandform]

Correct. Now do the one I just posted, and I'll give you a cookie. I dont know how to do it other than graphing it and then proving it via synthetic division =(. I can gives the answer but I don't know the math behind it, all I can do is graph it and then work backwards. (synthetic division) Spoiler for ok: X^3 means 3 answers. x = -3, x = -2, x = -1 and then something else with the >< stuff. Crap I made a mistake. I forgot to divide the 7 by 2. Fixed it. It should only be 2. You should end up with X = 2 and x = - 9... X = 9 makes the solution 5 = -5. Since radicands can't be negative and (3*-9) - 2 = -29, it can't be -9.

[drewmandan]

I dont know how to do it other than graphing it and then proving it via synthetic division =(. I can gives the answer but I don't know the math behind it, all I can do is graph it and then work backwards. (synthetic division) Spoiler for ok: X^3 means 3 answers. x = -3, x = -2, x = -1 Use the factor theorem. And I said greater than or equal, dammit! It should only be 2. You should end up with X = 2 and x = - 9... X = 9 makes the solution 5 = -5. Since radicands can't be negative and (3*-9) - 2 = -29, it can't be -9. Hmm quite right. I can see how the one answer isn't applicable, but I'm trying to figure out why my math didn't reveal that...

[Sandform]

x^4-15x^2+36

[drewmandan]

x^4-15x^2+36 Trivial. Use the substitution u = x^2 to reduce to u^2 -15u +36. Then u=-12 and -3 (assuming you want roots). Then the roots of x are ±sqrt(12)i, ±sqrt(3)i.

[Sandform]

Trivial. Use the substitution u = x^2 to reduce to u^2 -15u +36. Then u=-12 and -3 (assuming you want roots). Then the roots of x are ±sqrt(12)i, ±sqrt(3)i. Pfft, WRONG!. j/k, j/k, j/k, not really, j/k, j/k.

[Sandform]

Trivial. Use the substitution u = x^2 to reduce to u^2 -15u +36. Then u=-12 and -3 (assuming you want roots). Then the roots of x are ±sqrt(12)i, ±sqrt(3)i. Pfft, WRONG!. j/k, j/k, j/k, not really, j/k, j/k. I actually was serious about the answer being wrong though... there doesn't need to be an "i" and the sqrt(12) should be reduced to 2Sqrt(3) 2x + 4b = 0 4X + 2b = 2

[drewmandan]

I actually was serious about the answer being wrong though... there doesn't need to be an "i" and the sqrt(12) should be reduced to 2Sqrt(3) 2x + 4b = 0 4X + 2b = 2 Oh another stupid mistake. Yes, you're right. The roots are real. And fuck off with your lowest terms crap. No one cares about lowest terms past high school. As for that problem, I'm not sure what you want me to do. Is b a constant?