Thread: Equations

3(9x)/4x = 5x+4.5

18x = 20x^2 + 18x
20x^2 = 0
x = 0

Code:

Solve for X:
x^4 + 7x^3 + 17x^2 + 29x + 10

You can actually solve that with the basic rules. No logarithms, even!

TLS, I think you might be mistaken... 3*9 ≠ 18, and x=0 in 3(9x)/4x = 5x+4.5 is undefined. I got

Code:

3(9x)/4x = 5x+4.5
27x/4x = 5x+4.5
6.75 = 5x+4.5
5x = 2.25
x = 0.45

To keep things moving though, I'll do yours

Code:

x^4 + 7x^3 + 17x^2 + 29x + 10 = 0

probably can be factored into the form
(x^2 + ax + b)(x^2 + cx + d) = 0

x^4 + cx^3 + dx^2
    + ax^3 +acx^2 + adx
           + bx^2 + bcx + bd

x^4 + (c+a)x^3 + (d+ac+b)x^2 + (ad+bc)x + bd

c+a = 7
d+ac+b = 17
ad+bc = 29
bd = 10

by the almighty "guess & check" method
a=2,b=5,c=5,d=2

(x^2 + 2x + 5)(x^2 + 5x + 2) = 0

wah, quadratic formula T_T
(x^2 + 2x + 5) ->
x = (-2 ± sqrt(2^2 - 4*1*5))/2*1
x = -1 ± .5sqrt(-16) -> no real solutions :(

next one looks promising though
(x^2 + 5x + 2) ->
x = (-5 ± sqrt(5^2 - 4*1*2))/2*1

>>> x = -2.5 ± .5sqrt(17) <<<

Seems to work in the original equation. Why the weird answer?

Here's mine:

Code:

Solve for x and y, where x and y are integers and x>y:
sqrt(17+sqrt(264)) = sqrt(x) + sqrt(y)

You can solve this with basic algebra too

^arnt you suppose to have two equations for that? unless you meant to say x>y by one or two or something

Originally Posted by tkdyo

^arnt you suppose to have two equations for that? unless you meant to say x>y by one or two or something

It's enough to specify that x and y are integers. You're right that without that specification it would be a function.

Originally Posted by tkdyo

^arnt you suppose to have two equations for that? unless you meant to say x>y by one or two or something

It is a graphed equation. a curve/line. A system of equations is different. If he said:

3x+2y=2
7x-5y=12

Then yes, he could solve for both y and x. But with only one equation, one must graph it.

I know how systems of equations work, Ive done em toooo much, lol. but I forgot you said we could do graphical problems too.

Algebrahelp.com has a graphing calc in case we need it.

Let me rephrase: there is only one set of integers (x,y) such that sqrt(17+sqrt(264)) = sqrt(x) + sqrt(y). If x or y is any other value, there is a y or x that satisfies the equation, but one of them will not be an integer.

This problem can be solved entirely analytically using pretty basic algebra and a little mental arithmetic, but hey, use a graphing calculator if you must.

Algebra, eh? Ok:

Prove that are no simple groups of order 36.

gah, its only annoying me because I forgot what really changes things if they are integers. Ive been all the way through calc 4 with relative ease but I forgot basic stuff like this lol. closest Ive come is x =19, y = 2?

edit: drew; 36 is divisible by 3 which is a prime number other than 1...so a simple group exists?

the difference between sqrt(x) and sqrt(y) is approx. 5,766

I give up. I can't get any further than that. Explain please

Oops, I edited my post.

y = 5.766(5.766 - 2*sqrt(x)) + x

Am I right?

Originally Posted by aioinae

Let me rephrase: there is only one set of integers (x,y) such that sqrt(17+sqrt(264)) = sqrt(x) + sqrt(y). If x or y is any other value, there is a y or x that satisfies the equation, but one of them will not be an integer.

This problem can be solved entirely analytically using pretty basic algebra and a little mental arithmetic, but hey, use a graphing calculator if you must.

17+sqrt(264) = x + y + 2sqrt(xy)
17+sqrt(4*66) = x+y + sqrt(4*xy)

So we should find an (x,y) so that x+y = 17 and xy=66. Therefore, one must equal 6 and the other 11. In other words, either (6,11) or (11,6) will work.

That is about right, but I've checked it only mentally. Err... Where's your equation? O.o

here's one

Code:

Give the radius and center coordinates of this circle:
2((x+3)^2 - x - 102.5) = -((y+3)^2 - 187)

Originally Posted by ThreeLetterSyndrom

Code:

Give the radius and center coordinates of this circle:
2((x+3)^2 - x - 102.5) = -((y+3)^2 - 187)

Are you sure that's a circle? This is what I got for a plot, more of an arc:


Oh and by the way, gj drewmandan See, wasn't that easy?

I derived the formula from a circle

If you graph it, you should have negative numbers in it too!

He's right, you only took the positive square roots.

Solve for X

Code:

(sin x)^2 + (cos x )^2 = x + 5(i^2) - 1

It's simpler than it looks.

Ah, I see. I zoomed out and it's... still not a circle. That's an ellipse.

Code:

2((x+3)^2 - x - 102.5) = -((y+3)^2 - 187)
2(x^2 + 6x + 9 - x - 102.5) = -(y+3)^2 + 187
2(x^2 + 5x - 93.5) = -(y+3)^2 + 187
complete the square:
(5/2)^2 = 6.25
6.25-(-93.5) = 99.75
99.75 * 2 = 199.5
2(x+2.5)^2 + (y+3)^2 = 386.5
((x+2.5)^2) / 193.25 + ((y+3)^2) / 386.5 = 1

An ellipse centered at (-2.5,-3), whose radius ranges from about 13.901 (sqrt 193.25) horizontally to about 19.660 (sqrt 386.5) vertically.

Hang on a min, I think I made some mistake...

oh crap, I did make a mistake >.<
Anyway, ninja:

Code:

sin(x)^2 + cos(x)^2 = x + 5*-1 - 1      [i^2 = -1]
1 = x - 6                                          [sin(x)^2 + cos(x)^2 = 1]
5 = x                                               [+ 6]
x = 5

All these questions so far have been disgusting.

(sin x)^2 + (cos x )^2 = x + 5(i^2) - 1

lawl... less of an equation to solve, more of a collection of facts.

I've got myself a new one. This one should be correct.

Code:

It's an ellipse. Give radius and center coordinates.
x + 4/x + 3 = -y - 2.25/y - 3

Originally Posted by Xei

All these questions so far have been disgusting.

Give us an equation that someone in high school Algebra could solve, then. I'd like to see what's not "disgusting" according to Your Highness.

1 + 6 = 7, by the way.

Give us an equation that someone in high school Algebra could solve, then. I'd like to see what's not "disgusting" according to Your Highness.

Don't worry, I wasn't being patronising. It's just that the questions so far have been messy; just applications of formulae without anything very interesting about them and using messy numbers, etcetera.