1. ## Simplifying Equations

Hello,

I have two questions! I dont' want to be spoon fed the answers :-) just need some direction on websites, and, the categories of maths that these questions relate to if Im not using the right terms.

1. Is it possible to simplify this equation; z = ((x/2)+0.5)*x ?

If it is are there any web pages you could direct me to so I can teach myself
what I need to know to simplify it?

2. Are there any web pages that can show me what I need to know to re-arrange this equation so I get x from z?

It's been a while since I have done much with maths, so please be patient with me :-)

2. Originally Posted by rsmithers.net
I have two questions! I dont' want to be spoon fed the answers :-) just need some direction on websites, and, the categories of maths that these questions relate to if Im not using the right terms.

1. Is it possible to simplify this equation; z = ((x/2)+0.5)*x ?

If it is are there any web pages you could direct me to so I can teach myself
what I need to know to simplify it?

2. Are there any web pages that can show me what I need to know to re-arrange this equation so I get x from z?
Welcome to MHF Smithers! A good website to goto for algebra help is Algebra.help -- Calculators, Lessons, and Worksheets actually!

Your equation can easily be simplified. You just need to multiply your terms out:

$z = ((x/2)+0.5)*x$

$z = x(x/2)+0.5x$

$z = \frac{x^2}{2} + \frac{x}{2}$

You may go a step further if you choose to group the fractions together...

$z = \frac{x^2+x}{2}$

3. Originally Posted by colby2152
Welcome to MHF Smithers! A good website to goto for algebra help is Algebra.help -- Calculators, Lessons, and Worksheets actually!

Your equation can easily be simplified. You just need to multiply your terms out:

$z = ((x/2)+0.5)*x$

$z = x(x/2)+0.5x$

$z = \frac{x^2}{2} + \frac{x}{2}$

You may go a step further if you choose to group the fractions together...

$z = \frac{x^2+x}{2}$
Thanks, I understand some of what you have done, the ^2 part I shall have to work on. The simplified equation does of course work, nice one :-)

I should now find it easier (not easy) to work out how to re-arrange to get x from z, some of this maths is comming back to me slowly, but it was over 15 years ago!

Merry Christmas

4. Originally Posted by rsmithers.net
Thanks, I understand some of what you have done, the ^2 part I shall have to work on. The simplified equation does of course work, nice one :-)

I should now find it easier (not easy) to work out how to re-arrange to get x from z, some of this maths is comming back to me slowly, but it was over 15 years ago!

Merry Christmas
$x^2 = x*x$

Think of it like this:

$x^nx^m = x^{n+m}$

5. Originally Posted by rsmithers.net
Thanks, I understand some of what you have done, the ^2 part I shall have to work on. The simplified equation does of course work, nice one :-)

I should now find it easier (not easy) to work out how to re-arrange to get x from z, some of this maths is comming back to me slowly, but it was over 15 years ago!

Merry Christmas
I think I get it the powers aspect.

Is it because x times half x is the same as (x * x)/2 which is (x^2)/2?

6. Originally Posted by rsmithers.net
I think I get it the powers aspect.

Is it because x times half x is the same as (x * x)/2 which is (x^2)/2?
Yes... now think about it with numbers. Let x = 4. Half of x is now 2. Four times two is eight. Half of four squared is also eight. The equality checks out. There are dozens of ways you can relate any type of math problem. You just need to find what works best for you.

7. Originally Posted by colby2152
Yes... now think about it with numbers. Let x = 4. Half of x is now 2. Four times two is eight. Half of four squared is also eight. The equality checks out. There are dozens of ways you can relate any type of math problem. You just need to find what works best for you.
Thanks.

I know I have got the rearrangement wrong but so far this is what I have;

Example; (10^2 + 10) /2 = 55
so to get 10 from 55 I tried; x = 2z - (2 * sqrt(z))
like; 2*55 - 14.83239697

Clearly im wrong, any pointers?

8. Actually, using the quadratic formula, you should get this for x:

$x = \frac{-1 \pm \sqrt{1+8z}}{2}$

9. Originally Posted by colby2152
Actually, using the quadratic formula, you should get this for x:

$x = \frac{-1 \pm \sqrt{1+8z}}{2}$
Quick question, to the right of the -1 you have what seems like a plus minus symbol together, or a + with an underline, what does that mean?

Many thanks!

10. Originally Posted by rsmithers.net
Quick question, to the right of the -1 you have what seems like a plus minus symbol together, or a + with an underline, what does that mean?

Many thanks!
Plus or minus, there are two solutions for x.

$ax^2+bx+c=0$

To solve for this basic quadratic, use the following formula:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

The quadratic is of degree two, so it will have at most two points where it crosses or hits the x-axis.