# *HeLp* geometry & algebra question

• Feb 11th 2008, 06:31 PM
Afterme
*HeLp* geometry & algebra question
Hey guys I need help on one last problem well, actually two.

1. A rectangular parcel of land is 70 feet longer then it is wide. Each diagonal between opposite corners is 130ft. What are the dimensions of the parcel?

I know the L = 70 + W but other then that, no idea how to even start on this problem.

2. For what values of x is the expression Sqroot of 6x-x^2 defined as a real number?

this one I'm just at a pure loss.

thanks once again (Hi)
• Feb 11th 2008, 07:14 PM
topher0805
1) You know that $\displaystyle L = W + 70$, and you know that from corner to corner the length is 130 feet. You need to use pythagorean theory to solve this one. The hypotenuse is 130, and the other sides are W, and W + 70.

This gives us:

$\displaystyle 130^2 = W^2 + (W + 70)^2$

Now we simply have to solve for W. First multiply out the last term:

$\displaystyle 130^2 = W^2 + (W^2 + 140W + 4,900)$

Note that $\displaystyle 130^2$ is equal to $\displaystyle 16,900$. Now, bring everything to one side:

$\displaystyle 2W^2 + 140W - 12,000 = 0$

Now just use the quadratic equation to find W. :)

2) The square root function is only defined for numbers greater than or equal to 0. Therefore:

$\displaystyle 6x - x^2 >= 0$

Now just solve the inequality for x:

$\displaystyle 6x >= x^2$

So:

$\displaystyle 6 >= x$

We now have that $\displaystyle x$ must be equal to or less than 6 for the function to be defined.
• Feb 11th 2008, 08:47 PM
Afterme
2) The square root function is only defined for numbers greater than or equal to 0. Therefore:

$\displaystyle 6x - x^2 >= 0$

Now just solve the inequality for x:

$\displaystyle 6x >= x^2$

So:

$\displaystyle 6 >= x$

did you divide out the X?

We now have that $\displaystyle x$ must be equal to or less than 6 for the function to be defined.[/QUOTE]

ok i'm having a hard time understanding how to get from this
$\displaystyle 6x >= x^2$ to this
$\displaystyle 6 >= x$

if you can elaborate it would help alot thanks.
did you divide out the X?
• Feb 11th 2008, 08:50 PM
topher0805
I just divided both sides by $\displaystyle x$.

Oh, and I forgot to mention that $\displaystyle x$ must also be greater than 0.
• Feb 11th 2008, 08:58 PM
Afterme
Quote:

Originally Posted by topher0805
I just divided both sides by $\displaystyle x$.

Oh, and I forgot to mention that $\displaystyle x$ must also be greater than 0.

can you explain that a little bit. the X being greater than 0

hey thanks for all the help seriously (Hi)
• Feb 11th 2008, 09:08 PM
topher0805
When you divide both sides by $\displaystyle x$, there are actually two options. Either $\displaystyle x$ is positive or $\displaystyle x$ is negative.

In an inequality, when you divide by a negative number, you must switch the sign. So, the two cases are:

$\displaystyle x \leq 6$ when $\displaystyle x$ is positive.

and:

$\displaystyle x \geq 6$ when $\displaystyle x$ is negative.

You then test both solutions by plugging in values both above and below 6. By doing this you will notice that values above 6 make the function undefined. Therefore, $\displaystyle x$ can not be negative, and your final answer is:

$\displaystyle 0 \leq x \leq 6$
• Feb 11th 2008, 09:15 PM
Afterme
Thanks for helping me so much. Hopefully I'll get the knack of this stuff soon enough, until then I'll be counting on your help thanks!! (Hi)(Talking)
• Feb 11th 2008, 09:19 PM
topher0805
No problem. I struggle with math myself so any help I can provide I always do. I know what it feels like to be frustrated with a problem for a long time.

Just a reminder that there is a thanks button in the bottom right of posts to thank helpful posters. Just for future reference. :)
• Feb 11th 2008, 09:42 PM
angel.white
Quote:

Originally Posted by Afterme
can you explain that a little bit. the X being greater than 0

You have $\displaystyle \sqrt{6x-x^2}$

Consider if the value of 6x-x^2 were negative, lets say x = 7, then we have
$\displaystyle \sqrt{6(7)-7^2} = \sqrt{42-49} = \sqrt{-7}$

Lets say the answer of $\displaystyle \sqrt{-7}$ is equal to some value a.
Then
$\displaystyle \sqrt{-7}=a$
$\displaystyle -7=a^2$

What value can a have? It must be equal to -7 when multiplied by itself. But if a is positive, then a^2 is positive. And if a is negative, then a^2 is still positive (a negative number times a negative number is a positive number). So there is no real number which can be equal to a negative number when it is squared. This means that if $\displaystyle \sqrt{6x-x^2}$ is a real number, then 6x-x^2 must not be negative, so it must be greater than or equal to zero.