A coupla algebra problems

[psi20]

If kx^2 - kx - 6 is divisible by both (x + 1) and (x + m), find the value of m.

 

Given 2/x = y/3 = x/y , find x^3.

 

One way to pack a 100 by 100 square with 10,000 circles, each of diameter 1, is to put them in 100 rows with 100 circles in each row. If the circles are repacked so that the centers of any three tangent circles form an equilateral triangle, what is the maximum number of additional circles that can be packed?

[Primarygun]

First, find k , then factorize the given expression by (x+1). m then is revealed.

If find x^3 in terms of y, then just simply find x^2 and x.

Sorry, I don't understand the third one.

[psi20]

I know how to do the problems. These were for the reader to solve for fun.

[Algebracus]

The third problem: We have alternating rows of 100 and 99 circles, and the

vertical distance between the centres of circles in neighbouring rows is [MATH]\frac{\sqrt3}{2}[/MATH]. Let n be the number of rows. Then [MATH]1 + \frac{\sqrt3}{2}(n - 1) < 100 < 1 + \frac{\sqrt3}{2}n[/MATH], or [MATH]n - 1 < \frac{198}{\sqrt3} < n[/MATH], giving [MATH]n = 114[/MATH]. By this, we have 57 rows with 100 circles and 57 rows with 99 circles, totally 11343 circles, 1343 more circles than in the first packing.

[Ducky Havok]

For number 2, first you multiply each equation by 3, y, and x. That'll give

 

[math]6Y=Y^2X=3X^2[/math] By just looking at the 2nd part, you get

[math]Y^2X=3X^2[/math], and [math]Y^2=3X[/math] so [math] Y=\sqrt{3X}[/math]

so now you have the equation [math] 6Y=3X^2 [/math] after dividing by 3 and substituting in for y, you get

[math]2\sqrt{3X}=X^2 [/math] and when you solve for x,

[math]X=(2\sqrt{3})^\frac{2}{3}[/math] when you cube that you get

[math]X^3=12[/math]

[Ducky Havok]

For the first one, you can find K really easily. Plug in -1 for x and set it equal to 0 and you K=3.

Then you have [math] 3X^2-3X-6 [/math]

You already know one factor of this, so if you divide you get [math] (3X-6) [/math] as the other. Once you divide by 3 you get it in the form given above, so M=-2

[Asimov Pupil]

i thought it was 9604 for number 3

[cosine]

For the first one' date=' you can find K really easily. Plug in -1 for x and set it equal to 0 and you K=3.

Then you have [math'] 3X^2-3X-6 [/math]

You already know one factor of this, so if you divide you get [math] (3X-6) [/math] as the other. Once you divide by 3 you get it in the form given above, so M=-2

 

Don't you mean divide by [math](3X+6)[/math]? Then you find that m = 2.

[Dr. Zimski]

For the first one, I made an equation setting (kx^2 - kx - 6)/(x + 1) equal to (kx^2 - kx - 6)/(x + m).

 

You can solve it, but it's obvious by just looking at it that m = 1.

[Connor]

that doesn't seem like a good way to go about it.

 

The problem has two answers, so it's not necessarily one, and we're not trying to equate those sides, just factor one equation

[matt grime]

Nothing like resurrecting very old dead threads; check the dates, lads.