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Find two consecutive odd integers such that the sum of their squares is 130.

f(s) = 4s4 - 17s2 + 4

inequalities
t3 < 9t2

4x(x + 1) ≥ 3

polynomials
250x2 - 2x5

p4 - q4

finding all zeros
f(t) = (t - 3)2 - (t - 2)

2006-12-16 16:44:51 · 5 answers · asked by Nathalie d 1 in Science & Mathematics Mathematics

5 answers

"Find two consecutive odd integers such that the sum of their squares is 130."

Remember that consecutive odd integers differ from each other by 2. So if one of these integers is x, then the next consecutive odd integer would be x + 2.

Since the sum of their squares is 130, it's a matter of algebra:

(first number)^2 + (second number)^2 = 130
x^2 + (x + 2)^2 = 130

And we just expand,

x^2 + x^2 + 4x + 4 = 130
2x^2 + 4x + 4 = 130

Divide everything by 2, to get
x^2 + 2x + 2 = 65

And now, bring the 65 over to the left hand side, to get

x^2 + 2x - 63 = 0, which factors into
(x + 9) (x - 7) = 0, therefore x = -9 or x = 7, two solutions.

If x = -9, then x+2 = -7
If x = 7, then x = 9

Therefore, the two consecutive odd numbers are either {-9, -7} or {7, 9}

Inequality: t^3 < 9t^2

Our first step in solving this is to bring the 9t^2 over to the left hand side.

t^3 - 9t^2 < 0

Then, we factor t^2 out of the left hand side, to give us

t^2 [t - 9] < 0

And so our critical values are t = {0, 9}. Test the region AROUND those points by testing values less than 0, values between 0 and 9, and values greater than 9.
Test -1000000. t^2 [t - 9] becomes (positive) times (negative) = negative. So we include the interval (-infinity, 0).

Test 1: t^2 [t - 9] becomes (positive) times (negative) = negative. So we include (0, 9)

Test 1000000: t^2 [t - 9] becomes (positive) times (positive) = positive. So we do not include the interval (9, infinity).

Solution set of t: (-infinity, 0) U (0, 9)

"4x(x + 1) ≥ 3"

First off, expand the left hand side.

4x^2 + 4x >= 3

Bring the 3 over.

4x^2 + 4x - 3 >= 0, then factor,
(2x + 3 ) (2x - 1) >= 0
Our critical values are x = -3/2 and x = 1/2.

Test -1000000: (negative) times (negative) = positive => include interval.

Test 0: (positive) times (negative) = negative => exclude interval.

Test 10000000: (positive) times (positive) = positive => include interval.

Solution set: x in the interval (-infinity, -3/2] U [1/2, infinity)

Note the square brackets. This is because it is >=.


"250x^2 - 2x^5"

First, factor the biggest term out.

2x^2 [125 - x^3]

Now, factor the stuff in the brackets as a difference of cubes. Steps to do this:
(1) First set of brackets, take the cube root of each term. In our case, it would be (5 - x).
(2) Second set of brackets; look at the first set of the bracket, and
(a) "square the first" --> square the first value (5^2 = 25)
(b) "negative product" -> Multiply 5 and -x, and then take the negative (5 * -x = -5x, which has a negative of +5x)
(c) "square the last" -> square the last value ([-x]^2 = x^2.

2x^2 [125 - x^3]
2x^2 [5 - x] [25 + 5x + x^2]

"p^4 - q^4"

Factor as a difference of squares.
(p^2 - q^2) (p^2 + q^2)

Factor the first set also as a difference of squares.
(p - q) (p + q) (p^2 + q^2)

"Find the zeros of f(t) = (t - 3)^2 - (t - 2)"

No choice here but to expand and factor.

f(t) = t^2 - 6t + 9 - t + 2
f(t) = t^2 - 7t + 11
The "zeros" of f(t) are calculated by making f(t) equal to 0.

0 = t^2 - 7t + 11

We have to use the quadratic formula.

x = (-b +/- sqrt(b^2 - 4ac) ) / 2a
x = (7 +/- sqrt(49 - 4(1)(11)) / 2
x = (7 +/- sqrt(5))/2

Therefore, x = (7 + sqrt(5)/2 or x = (7 - sqrt(5)/2

2006-12-16 17:04:41 · answer #1 · answered by Puggy 7 · 1 0

f(s) = 4s4 - 17s2 + 4

inequalities
t3 < 9t2

4x(x + 1) ≥ 3
let the consecutive odd nos be n and n+2
n^2+(n+2)^2=130
n^2+n^2+4n+4=130
2n^2+4n-126=0
dividing by2
n^2+2n-63=0
(n+9)(n-7)=0
n=-9 or 7
so thenos are -9,-7 or 7 and 9
polynomials
250x2 - 2x5
dividing by 2x^2
2x^2(125-x^3)
2x^2(5-x)(25+5x+x^2)
p4 - q4
(p^2+q^2)(p^2-q^2)
(p^2+q^2)(p+q)(p-q)
finding all zeros
f(t) = (t - 3)2 - (t - 2)
(t-3)2-(t-2)
=2t-6-t-2
=t-8
so 8 is a zero of the polynomial

2006-12-16 16:55:46 · answer #2 · answered by raj 7 · 0 0

I do watch it and I thought the same thing I am glad I am not the only one that saw the resemblence because my daughter didnt think that they looked alike. I dont think that they have actually officially divorced yet but he went off to I believe medical school and she didnt wanna hold him back from his dreams now that he is better. So they are seperated perhaps. But yes that guy most definately resebles raj or if not is him. I am gonna try googling it and if I find out something I will edit my answer.

2016-05-23 01:23:22 · answer #3 · answered by Megan 4 · 0 0

Has it occurred to you that getting someone else to do your homework isn't going to do you one iota of good when you are faced with an exam?

2006-12-16 17:31:35 · answer #4 · answered by Anonymous · 0 0

look lady its the weekend which means party....if u wana keep on studying read ur math book to yourself an do every1 a favor ight later

2006-12-16 16:48:34 · answer #5 · answered by Anonymous · 0 2

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