Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#3 Re: Help Me ! » Irrational roots » 2015-01-23 17:58:22

yep thanks but please do not post solutions just help

#5 Help Me ! » Infintite sieries » 2015-01-19 17:48:40

Replies: 4

1) Let S be the sum of a finite geometric series with negative common ratio whose first and last terms are 1 and 4, respectively. (For example, one such series is 1-2+4, whose sum is 3.)

There is a real number L such that S must be greater than L, but we can make S as close as we wish to L by choosing the number of terms in the series appropriately. Determine L.

2)If  a/b rounded to the nearest trillionth is 0.008012018027, where a and b are positive integers, what is the smallest possible value of a+b?/

#6 Help Me ! » infinite arithmetic series » 2015-01-16 18:49:37

Replies: 6

heres the problem
(a)Determine all nonnegative integers r such that it is possible for an infinite arithmetic sequence to contain exactly r terms that are integers. Prove your answer.

seems like an awefull lot of cases

(b)Determine all nonnegative integers r such that it is possible for an infinite geometric sequence to contain exactly r terms that are integers. Prove your answer.

same thing

#7 Re: Help Me ! » Polynomial eugh » 2015-01-06 13:39:13

I found that 3i, -3i, and 2 work so theres one.
Why does one root have to be real when the coefficients are?

#8 Help Me ! » Polynomial eugh » 2015-01-05 17:30:28

Replies: 4

Let x, y, and z be complex (i.e., real or nonreal) numbers such that x+y+z, xy+xz+yz, and xyz are all positive real numbers.

Is it necessarily true that x, y, and z are all real, and positive? If so, prove it. If not, give a counterexample.

Thanks. I can prove that they are positive if they are real numbers, but don't see the approach for if they are complex.

#9 Re: Help Me ! » More of Vieta » 2014-12-14 15:04:29

The answer for q1 was 14. Interesting.
That means the poly was 2x^3+14x^2-82x+66.

q2 was 86 then.
Thank you for your help.

#10 Help Me ! » More of Vieta » 2014-12-12 21:06:26

Replies: 4

These are driving me mad:

(1)Suppose the polynomial

has integer coefficients, and its roots are distinct integers.

Given that a_n=2 and a_0=66, what is the least possible value of


(2)Two of the roots of the polynomial

have product -32.

Determine k.


be the roots of f(x).


#11 Re: Help Me ! » vieta's formulas » 2014-12-11 15:11:13

I just got another similar problem I thought was really cool:

Let r, s, and t be roots of the equation



rs/t+st/r+tr/s = ((rs+st+tr)^2-2rst(r+s+t))/rst = (6^2-2*9*(-5))/9 = -6

but strangly, -6 = -(rs+st+rt)...

I dont see how that works smile

I just thought that was a really cool factorization.

so that can also be

(b^2+2ac)/c = a

maybe I discovered something new. Duno smile

#12 Re: Help Me ! » vieta's formulas » 2014-12-10 19:58:15

@Agnishom, on your before last post, you said the answer to (a) is the coefficient of x divided by that of x^4. Thats seven, and its incorrect... I think the answer is -7

Thanks both of you

#13 Help Me ! » vieta's formulas » 2014-12-09 22:52:44

Replies: 8

For math class homework, I have several problems like this and I don't really know how to do these:

Let g(x) = x^4-5x^3+2x^2+7x-11, and let the roots of g(x) be p, q, r, and s.

(a) Compute pqr + pqs + prs + qrs.
(b) Compute

(c) Compute p^2 + q^2 + r^2 + s^2.
(d) Compute p^2qrs + pq^2rs + pqr^2s + pqrs^2.

Can you help me understand the method, not just do them for me?


#15 Help Me ! » Irrational roots » 2014-11-28 17:47:59

Replies: 8

(a) Give an example of two irrational numbers which, when added, produce a rational number. (sqrt(2) and -sqrt(2) big_smile)

Now let's consider just the addition of radicals.

(b) Suppose that a and b are positive integers such that both

are irrational. For what values of a and b is
rational? Prove your answer.

(c) Again assuming a and b positive integers such that both

are irrational, for what values of a and b is
rational? Prove your answer.

Thank you. I'm stumped!

#16 Re: Help Me ! » hard algebra » 2014-11-08 17:07:05

sorry, I was just really tired and not thinking.

#17 Help Me ! » hard algebra » 2014-11-08 00:05:38

Replies: 3

Let x=a and x=b be the roots of the equation x^2 - mx + 2 = 0.

Suppose that x=a+\frac 1b and x=b+\frac 1a are the roots of the equation x^2 - px + q = 0.

Determine the value of q.

#18 Help Me ! » Algebraic problemo » 2014-11-07 23:29:55

Replies: 1

Let a,b be real numbers such that 0<b <= a.

Prove that the equation x^2+ax-b=0 cannot have two integer roots.

#19 Re: Help Me ! » Hard algebra » 2014-11-01 20:02:37

thank you @Olinguito and @bob bundy

#20 Help Me ! » Hard algebra » 2014-10-31 23:00:33

Replies: 6

Let a,b be real numbers, and let x_1, x_2 be the roots of the quadratic equation x^2+ax+b=0.

Prove that if x_1, x_2 are real and nonzero,

, and b>0, then |a+2|>2.

#21 Re: Help Me ! » algebra » 2014-10-26 21:49:41

Thanks. Sorry I forgot to place the [math] tags...

#22 Help Me ! » algebra » 2014-10-25 22:27:16

Replies: 5

Prove that if w,z are complex numbers such that |w|=|z|=1 and wz\ne -1, then \frac{w+z}{1+wz} is a real number.

#23 Re: Help Me ! » some hard algebra problems » 2014-10-20 11:32:18

on problem 3 I took
(p+q+r)^2 = 7^2
p^2+q^2+r^2 = 9
and set
(p+q+r)^2-40 = p^2+q^2+r^2
(p^2+2 p q+2 p r+q^2+2 q r+r^2)-p^2-q^2-r^2 = 40
and 2 p q+2 p r+2 q r = 40
p q+p r+q r = 20,
but it askes for the average so the answer is 10, right?
well it says that is incorrect

#24 Re: Help Me ! » some hard algebra problems » 2014-10-20 11:20:18

Thank you guys so much!
These problems were completely beyond me...

#25 Help Me ! » some hard algebra problems » 2014-10-19 21:57:38

Replies: 10

1) Compute the sum

2) Find the ordered quintuplet (a,b,c,d,e) that satisfies the system of equations:

3) Suppose p+q+r = 7 and p^2+q^2+r^2 = 9. Then, what is the average (arithmetic mean) of the three products pq, qr, and rp?

4) Find the largest four-digit value of t such that

is an integer.

Board footer

Powered by FluxBB