Help solving: 16x - 8 = (110 + 2x^2) / 2

Richard

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Mar 5, 2010, 4:44:15 PM3/5/10

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Given:

Trapeziod

'Base 1' = 110
'Base 2' = 2x^2
Midsegment = 16x - 8

Solve for the length of the midsegment.

If anyone would be so kind as to demonstrate the correct way to solve this,
it would be greatly appreciately.

Thanks,
Richard

Brian M. Scott

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Mar 5, 2010, 5:13:51 PM3/5/10

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On Fri, 5 Mar 2010 13:44:15 -0800, Richard
<ev...@delete.com> wrote in
<news:NOWdnXGuQIms5gzW...@earthlink.com> in
alt.algebra.help:

> Given:

> Trapeziod

Typo: Trapezoid

> 'Base 1' = 110
> 'Base 2' = 2x^2
> Midsegment = 16x - 8

> Solve for the length of the midsegment.

Your subject line already has the right equation:

16x - 8 = (110 + 2x^2)/2

Multiply both sides by 2:

32x - 16 = 110 + 2x^2

Bring everything to one side and simplify:

0 = 126 - 32x + 2x^2
0 = 63 - 16x + x^2

To solve this, either use the quadratic formula, or, more
easily, factor it:

0 = (x - 7)(x - 9)
x - 7 = 0 or x - 9 = 0
x = 7 or x = 9
16x - 8 = 112 - 8 = 104 or 16x - 8 = 144 - 8 = 136

Either the base is 2 * 7^2 = 98 and the midsegment is 104,
or the base is 2 * 9^2 = 162, and the midsegment is 136.

[...]

Brian

Richard

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Mar 5, 2010, 5:37:29 PM3/5/10

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"Brian M. Scott" wrote:
> Richard wrote:
> >
> > Trapeziod
>
> Typo: Trapezoid

Thank you.

> 0 = (x - 7)(x - 9)
> x - 7 = 0 or x - 9 = 0
> x = 7 or x = 9
> 16x - 8 = 112 - 8 = 104 or 16x - 8 = 144 - 8 = 136
>
> Either the base is 2 * 7^2 = 98 and the midsegment is 104,
> or the base is 2 * 9^2 = 162, and the midsegment is 136.
>
> [...]
>
> Brian

Immensely appreciated. Thank you Brian.

Stan Brown

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Mar 5, 2010, 7:51:41 PM3/5/10

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Fri, 5 Mar 2010 17:13:51 -0500 from Brian M. Scott
<b.s...@csuohio.edu>:

> Your subject line already has the right equation:
>
> 16x - 8 = (110 + 2x^2)/2
>
> Multiply both sides by 2:
>
> 32x - 16 = 110 + 2x^2

Or save yourself some work and do the division that is already
indicated on the right-hand side:

16x - 8 = 55 + x^2

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai...

Richard

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Mar 5, 2010, 11:46:32 PM3/5/10

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Stan Brown wrote:

> Brian M. Scott wrote:
> >
> > 16x - 8 = (110 + 2x^2)/2
> >
> > Multiply both sides by 2:
> >
> > 32x - 16 = 110 + 2x^2
>
> Or save yourself some work and do the division that is already
> indicated on the right-hand side:
>
> 16x - 8 = 55 + x^2

Thanks Stan.

I was able to simplify it several ways, but for the life of me I could not
remember how to properly factor it.

Richard

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Mar 5, 2010, 11:55:42 PM3/5/10

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BTW: It sure is good to see a newsgroup still working and not filled with
nothing but garbage, long abandoned by anyone with any real substance to
offer.

Pfs...@aol.com

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Mar 6, 2010, 12:15:44 PM3/6/10

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Just give us your instructor's e-mail address and we'll send it
directly to him/her. We will of course tell the instructor that we did
it for you!
O,r as an alternative, study, learn, do the work yourself and don't
be a cheat!
Cheats are scum!

Richard

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Mar 6, 2010, 12:59:17 PM3/6/10

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Thank you for that kind advise. Since I have been out of school for some 30
years now, I am a little rusty currently. The kindness some here have been
gracious enough to show, has helped immensely in helping me to recall things
that I am sure are elementary to most here.

As for doing the work myself: I can assure you that with the exception of
that one problem, I have worked through the rest unassisted.

Again, thank you,
Richard

Barry Schwarz

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Mar 18, 2010, 11:47:25 PM3/18/10

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Factoring may not always be intuitive. The quadratic formula will
always work and is not that much more effort.

--
Remove del for email

Frederick Williams

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Mar 19, 2010, 1:34:19 PM3/19/10

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Barry Schwarz wrote:

> >Stan Brown wrote:

> >>
> >> 16x - 8 = 55 + x^2

> >
> Factoring may not always be intuitive. The quadratic formula will
> always work and is not that much more effort.

In this case it is

0 = 63 - 16x + x^2

that is to be solved, so the op needs to ask: what numbers multiply to
give 63 and add to give -16? Since the factors of 63 are

1 and 63
3 and 21
7 and 9

and those negated, it is not a difficult question. The problem with the
quadratic formula is that students love formulae into which they can
plug numbers, do a bit of manipulation, and have the answer pop out
_without any thought whatsoever_.

--
I can't go on, I'll go on.

Cobra Commander

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Jul 5, 2010, 7:51:19 PM7/5/10

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In article <NOWdnXGuQIms5gzW...@earthlink.com>,
"Richard" <ev...@delete.com> wrote:

Solution:

16x - 8 = (110+2x^2)/2

Carry the 2 to the left

32x - 16 = 110 + 2x^2

Bring all terms to the right, yo now have a quadratic

2x^2 -32x + 126 = 0

Normalize with respect to 2

x^2 - 16x + 63 = 0

Solve the quadratic

x = {16 +- (16^2-4*1*63)^.5}/(2*1)

x = (16 +- 2)/2

1st root, x = 18/2 = 9
2nd root, x = 14/2 = 7

Stan Brown

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Jul 5, 2010, 8:55:26 PM7/5/10

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On Mon, 05 Jul 2010 23:51:19 GMT, Cobra Commander wrote:

> Solution:

I don't disagree with your solution, but at two points you chose a
harder way instead of an easier way of doing things.

> 16x - 8 = (110+2x^2)/2
>
> Carry the 2 to the left

I *guess* you mean "multiply both sides by 2":

> 32x - 16 = 110 + 2x^2

But it's easier instead to do the indicated division on the right:

16x - 8 = 55 + x^2

Always try to keep numbers from getting any larger than necessary.
In particular, quadratics are easier to solve when the coefficient of
x^2 is 1.

> Bring all terms to the right, yo now have a quadratic
>
> 2x^2 -32x + 126 = 0

0 = x^2 - 16x + 63

> Solve the quadratic
> x = {16 +- (16^2-4*1*63)^.5}/(2*1)
> x = (16 +- 2)/2
> 1st root, x = 18/2 = 9
> 2nd root, x = 14/2 = 7

Yes, but this one is *very* easy to factor. You need two numbers
that multiply to 63; and because you have +63 and -16 you know that
the two numbers must both be negative. The solution jumps out:

0 = (x-7)(x-9) ==> x = 7 or 9