Quaratic Application Problem

[papaesmith]

I'm stuck and need help! I have spent three long hours trying to figure out how they reduced a certain part of Quadratic Equation application problem. This is a part I must reduce before I can put the equation into Quadratic Formula Form. Once I get past this, the rest will not be a problem.

Here is the part of the problem:

X[(110/Xsquared)/600]

...and it reduces to:

121/6X

I can't firue out how it reduces to 121/6X. This part was derived from the formula where:

The cost of fuel for a van is C=Vsquared/600 where V is the speed in miles per hour. The driver is paid $5.00 per hour. Find the speed if the cost for wages and fuel for a 110-mile trip is $20.39.

Can you help me?

Thanks,
Ed Smith

 

[stapel]

It would really help if you showed all the steps between the problem statement and "x [ (100/x²) / 600]". It seems to me highly likely that, since this expression does not "simplify" to 121/(6x), the error(s) occurred earlier.

Thank you.

Eliz.

 

[Denis]

X[(110/Xsquared)/600]
...and it reduces to:
121/6X

Hey papa, that reduces to 11/(60x)

Are you sure there's no typo in: X[(110/Xsquared)/600] ?

 

[stapel]

Note to tutors: For some reason, the poster provided the requested information in a private message:

The Math problem goes like this:

A small business uses a minivan to make deliveries. The cost per hour for fuel for the van is C = v^2/600, where v is the speed in miles per hour. The driver is paid $5.00 per hour. Find the speed if the cost for wages and fuel for a 110-mile trip is $23.39.

The solutions guide sets it up like this:

Total Cost = Wage Cost + Fuel Cost
Time = X
20.39 = 5X + X[(110/X^2)/600]

...which simplifies to 20.39 = 5X + 121/6X, and further reduces to 122.34X = 30X^2 +121, and then from there on into the quadratic formula which I can't write on a keyboard.

The answers are v = 110/20.39 = 46 miles/hour or v = 110/1.69 = 65 miles/hour.

See if you can figure out how X[(110/X^2)/600], simplifies to 121/6X. Nobody has been able to figure this part out, not even an online algebra tutor. Is the book wrong?

In particular, we see that the "110/x^2" should almost certainly be "(110/x)²", which causes the entire expression to simplify correctly.

Eliz.

 

[morson]

Confusing.

 

[jwpaine]

In mathematics, the fundamental theorem of algebra states that every non-zero single-variable polynomial, with complex coefficients, has exactly as many complex roots as its degree

But..... I always thought a "complex" root was one with an imaginary solution...with no x intercepts....

real solutions can be complex??

 

[morson]

In mathematics, the fundamental theorem of algebra states that every non-zero single-variable polynomial, with complex coefficients, has exactly as many complex roots as its degree

But..... I always thought a "complex" root was one with an imaginary solution...with no x intercepts....

real solutions can be complex??

The fundamental theorem of algebra can be questioned when you look at an equation like x^2 + 6x + 9 = 0. You'd normally say there's only 1 solution (x = -3), but because the quadratic occurs in two separate linear brackets (x + 3)(x + 3), they're seen as two different solutions, called roots of multiplicity.

And yeah, real solutions are complex numbers of the form a + 0i, where a is real.

 

[Mrspi]

[quote="jwpaine]
But..... I always thought a "complex" root was one with an imaginary solution...with no x intercepts....

real solutions can be complex??[/quote]

Yes....every real number is also a complex number.

For example,

2 = 2 + 0i
sqrt(5) = sqrt(5) + 0i
pi = pi + 0i

Every real number is also a complex number.

 

[Denis]

I think this is the most confusing thread I've seen at this site!!

Apparently, the original problem is:
A small business uses a minivan to make deliveries.
The cost per hour for fuel for the van is C = v^2/600, where v is the speed in miles per hour.
The driver is paid $5.00 per hour.
Find the speed if the cost for wages and fuel for a 110-mile trip is $23.39.

$23.39 is a typo? Should be 20.39, apparently.

The time in hours = 110 / v : why use X?!

Cost for 1 hour = v^2 / 600 + 5

So cost for 110 / v hours = (110 / v)(v^2 / 600 + 5) = 20.39
which simplifies to:
110v^2 - 12234v + 330000 = 0
solve to get v = ~65.220 or v = ~45.998

 

[stapel]

I think this is the most confusing thread I've seen at this site!!

For some reason, it appears that the tutors carried their discussion from another thread into this one...?

why use X?

Dunno, but that's what the solution manual used, which is what produced the expression in "x" rather than, as you and I would have prefferred, in "v". But, as was pointed out before the digression, the problem was not with the variable, but with the grouping symbols -- or, rather, the lack thereof.

Eliz.

 

[Denis]

Yes, you're correct Stapel: all due to incorrect grouping!

> The solutions guide sets it up like this:
> Total Cost = Wage Cost + Fuel Cost
> Time = X
> 20.39 = 5X + X[(110/X^2)/600]

Since time = X, then v = 110/X, so that last term should be:
X[(110/X)^2 / 600] : v^2 / 600, as stated in problem.

> ...which simplifies to 20.39 = 5X + 121/6X, and further reduces to 122.34X = 30X^2 +121,
> and then from there on into the quadratic formula which I can't write on a keyboard.

So above is correct; quadratic equation is:
30X^2 - 122.34X + 121 = 0
Solve to get X = ~2.39 or X = ~1.69

> The answers are v = 46 miles/hour or v = 65 miles/hour.

That's now correct; since v = 110 / X, then:
v = 110 / ~2.39 = ~46 or v = 110 / ~1.69 = ~65

PS: the "teacher" that specified use of "X = time" should be shot at sunrise :shock:

 

[stapel]

PS: the "teacher" that specified use of "X = time" should be shot at sunrise :shock:

It seems that, many times, the solution manuals aren't written by the authors, but are instead composed by graduate students. Sleep-deprived, under-fed, over-caffienated graduate students.

Sometimes weird stuff ensues... :wink:

Eliz.