How can I solve equations without graphing?

[repugno]

Can't seem to get my head round this question. Can anyone help please

A rectangular box with no lid is made from thin cardboard. The base is 2x centimetres long and x centimetres wide and the volume is 48 cubic centimetres. Show that the area, y square centimetres, of cardboard used is given by y = 2x^2 + 144x^-1

thank you

 

[jamesrc]

The area of the base is 2x², correct? (length times width).

There are 4 other sides to the box (since there's no lid). Two of those sides have a base of 2x centimeters, while the other 2 have a base of x centimeters (draw the picture if that is not clear to you. All of the sides have the same height. To find the height, divide the volume by the area of the base: 48/(2x*x).

Put it all together:

A_s = 2x^2 + 2\left(2x\frac{48}{2x^2} + x\frac{48}{2x^2} \right)

where A_s is the surface area of cardboard used. Make sense?

 

[repugno]

Makes pleasant sense now, thanks

 

[repugno]

Hello all,
I’m having some trouble solving equations, perhaps someone could explain to me how I could do so apart from plotting a graph.

Eg. Normally I would do this ...

0 = x^3 + 3x^2 - 4
4 = x^3 + 3x^2
4 =(x^2 + 3x)x

So, x = 4

0 =(x + 3)x

So, x = 0

0 =(x + 3)

So, x =-3

This is clearly not the correct method

Thank you.

 

[deltabourne]

Originally posted by repugno
Hello all,
I’m having some trouble solving equations, perhaps someone could explain to me how I could do so apart from plotting a graph.

Eg. Normally I would do this ...

0 = x^3 + 3x^2 - 4
4 = x^3 + 3x^2
4 =(x^2 + 3x)x

So, x = 4

0 =(x + 3)x

So, x = 0

0 =(x + 3)

So, x =-3

This is clearly not the correct method

Thank you.

You went wrong at (x^2+3x)*x=4. That does not imply x=4. We can do something similar when we have something like a*b=0, since we know that one of the products (a or b) must be 0.

 

[Moonbear]

Originally posted by repugno
Hello all,
I’m having some trouble solving equations, perhaps someone could explain to me how I could do so apart from plotting a graph.

Eg. Normally I would do this ...

0 = x^3 + 3x^2 - 4
4 = x^3 + 3x^2
4 =(x^2 + 3x)x

So, x = 4

0 =(x + 3)x

So, x = 0

0 =(x + 3)

So, x =-3

This is clearly not the correct method

Thank you.

There's something amiss with the original equation. If you are graphing it, you need Y in there somewhere to have two coordinates to plot. There is then no unique solution to solve for X, it is an equation describing how one variable, X, changes in relation to another variable, Y, for any X you choose. With the above equation setting Y as 0, there is still no unique solution. Both 1 and -2 would be possible values of X, but I can't explain that as anything other than trial and error. If you're still solving these equations by graphing them in your class, stick with that. The reason for it is so you *see* the relationships with different types of equations. If you continue taking higher level math classes, more will be revealed then, or perhaps even later this year in your current class. In math, everything is taught very sequentially so you learn now what you will need to understand next year's class, and so on. It's not a good idea to try jumping too far ahead or you'll miss important concepts along the way.