am I on the right track here?

[JeffM]

but, if I use the ratio .5/0.3333=x/1.5 I get 2.25 and this IS a good approximation of the right answer. .................edited

There are certainly problems for which only an approximate answer is possible, and such problems are studied in numerical methods. But those methods require a specification of when an approximation is good enough. In this case, you can at least try for an exact answer.

There are only six possible rational roots to try according to the rational root theorem, and the negative ones do not look promising.

[MATH]8 \left ( \dfrac{1}{8} \right )^3 - 18 \left ( \dfrac{1}{8} \right )^2 + 1 = \dfrac{1}{64} - \dfrac{18}{64} + \dfrac{64}{64} \ne 0.[/MATH]
[MATH]8 \left ( \dfrac{1}{4} \right )^3 - 18 \left ( \dfrac{1}{4} \right )^2 + 1 = \dfrac{1}{8} - \dfrac{18}{16} + \dfrac{8}{8} = \dfrac{1 + 8 - 9}{8} = 0.[/MATH]
[MATH]\therefore r = \dfrac{1}{4} \text { is one solution.}[/MATH]
[MATH]\dfrac{8r^3 - 18r^2 + 1}{r - \dfrac{1}{4}} = \dfrac{32r^3 - 72r^2 + 4}{4r - 1} = \dfrac{32r^3 - 8r^2 + 8r^2 - 72r^2 + 4}{4r - 1} =[/MATH]
[MATH]8r^2 - \dfrac{64r^2 - 4}{4r - 1} = 8r^2 - \dfrac{64r^2 - 16r + 16r - 4}{4r - 1} = 8r^2 - 16r - \dfrac{16r - 4}{4r - 1} = 8r^2 - 16r - 4.[/MATH]
[MATH]\left (r - \dfrac{1}{4} \right ) (8r^2 - 16r - 4) = 8r^3 - 16r^2 - 4r - 2r^2 + 4r + 1 = 8r^3 - 18r^2 + 1.[/MATH]
[MATH]16^2 - 4(8)(-\ 4) = 16(16 - 8) = 8 * 2 * 8(2 - 1) = 8^2 * 2 * 1 = 8^2 * 2.[/MATH]
[MATH]r = \dfrac{16 \pm \sqrt{8^2 * 2} }{2 * 8} = \dfrac{16 \pm 8\sqrt{2}}{16} = 1 \pm 0.5 \sqrt{2}.[/MATH]
At this point, you have three EXACT answers. The two involving the square root of 2 can be given a decimal approximation to any required degree of accuracy.

You do not need anything but algebra and arithmetic.

 

[Deleted member 4993]

There are certainly problems for which only an approximate answer is possible, and such problems are studied in numerical methods. But those methods require a specification of when an approximation is good enough. In this case, you can at least try for an exact answer.

There are only six possible rational roots to try according to the rational root theorem, and the negative ones do not look promising.

[MATH]8 \left ( \dfrac{1}{8} \right )^3 - 18 \left ( \dfrac{1}{8} \right )^2 + 1 = \dfrac{1}{64} - \dfrac{18}{64} + \dfrac{64}{64} \ne 0.[/MATH]
[MATH]8 \left ( \dfrac{1}{4} \right )^3 - 18 \left ( \dfrac{1}{4} \right )^2 + 1 = \dfrac{1}{8} - \dfrac{18}{16} + \dfrac{8}{8} = \dfrac{1 + 8 - 9}{8} = 0.[/MATH]
[MATH]\therefore r = \dfrac{1}{4} \text { is one solution.}[/MATH]
[MATH]\dfrac{8r^3 - 18r^2 + 1}{r - \dfrac{1}{4}} = \dfrac{32r^3 - 72r^2 + 4}{4r - 1} = \dfrac{32r^3 - 8r^2 + 8r^2 - 72r^2 + 4}{4r - 1} =[/MATH]
[MATH]8r^2 - \dfrac{64r^2 - 4}{4r - 1} = 8r^2 - \dfrac{64r^2 - 16r + 16r - 4}{4r - 1} = 8r^2 - 16r - \dfrac{16r - 4}{4r - 1} = 8r^2 - 16r - 4.[/MATH]
[MATH]\left (r - \dfrac{1}{4} \right ) (8r^2 - 16r - 4) = 8r^3 - 16r^2 - 4r - 2r^2 + 4r + 1 = 8r^3 - 18r^2 + 1.[/MATH]
[MATH]16^2 - 4(8)(-\ 4) = 16(16 - 8) = 8 * 2 * 8(2 - 1) = 8^2 * 2 * 1 = 8^2 * 2.[/MATH]
[MATH]r = \dfrac{16 \pm \sqrt{8^2 * 2} }{2 * 8} = \dfrac{16 \pm 8\sqrt{2}}{16} = 1 \pm 0.5 \sqrt{2}.[/MATH]
At this point, you have three EXACT answers. The two involving the square root of 2 can be given a decimal approximation to any required degree of accuracy.

You do not need anything but algebra and arithmetic.

Small arithmetic error (4th line from bottom). Correcting it, we get:

[MATH]r = 1 \pm \sqrt{\dfrac{3}{2}}.[/MATH]
All other interpretations remain valid.

 

[JeffM]

A sign

Small arithmetic error (4th line from bottom). Correcting it, we get:

[MATH]r = 1 \pm \sqrt{\dfrac{3}{2}}.[/MATH]
All other interpretations remain valid.

A sign error! LOL. Off to the corner with Jeff. AGAIN.

[MATH]16^2 - 4(8)(-\ 4) = 16^2 + 16 * 8 = 16(16 + 8) = 16(8)(2 + 1) = 2 * 8 * 8 * 3 = 6 * 8^2 \implies[/MATH]
[MATH]\dfrac{16 \pm \sqrt{6 * 8^2}}{2 * 8} = 1 \pm \sqrt{\dfrac{6 * 8^2}{16^2}} = 1 \pm \sqrt{\dfrac{3 * 2^7}{2^8}} = 1 \pm \sqrt{\dfrac{3}{2}}.[/MATH]

 

[allegansveritatem]

The function (derivative) in question does NOT have a rational root at 2.25. By numerical approximation (Newton's method) we get x ~ 2.224744873 ~ 20/9 (not exact) as one of the roots......... edited

the roots are -,2, .25, 2.2. What is "numerical approximation"?

 

[allegansveritatem]

So I went back at this today using the rational root theorem and below is the result:

 

[Dr.Peterson]

That's just right; this is how I solved it. I'm certainly curious how you were expected to do it without knowing the Rational Root Theorem. If the course uses graphing calculators, you would probably use that. (By the way, their root finders use a numerical approximation method.)

I'd just give the answer as either exactly 1/4, 1 + sqrt(6)/2, and 1 - sqrt(6)/2; or approximately as 0.25, 2.225, and -0.225, giving more decimal places than you did for the approximate roots.

 

[allegansveritatem]

That's just right; this is how I solved it. I'm certainly curious how you were expected to do it without knowing the Rational Root Theorem. If the course uses graphing calculators, you would probably use that. (By the way, their root finders use a numerical approximation method.)

I'd just give the answer as either exactly 1/4, 1 + sqrt(6)/2, and 1 - sqrt(6)/2; or approximately as 0.25, 2.225, and -0.225, giving more decimal places than you did for the approximate roots.

The book does give a lot of attention to calculator use but most of the exercises do not have the calculator icon alongside them so they emphasis is on understanding the process rather than just getting a result. I'm not sure why the Rational Root Theorem has not been covered...could be he covered than in another book that he assumes this book's reader have already read.
Right, I will try to be more precise with my answers. Thanks

 

[allegansveritatem]

There are certainly problems for which only an approximate answer is possible, and such problems are studied in numerical methods. But those methods require a specification of when an approximation is good enough. In this case, you can at least try for an exact answer.

There are only six possible rational roots to try according to the rational root theorem, and the negative ones do not look promising.

[MATH]8 \left ( \dfrac{1}{8} \right )^3 - 18 \left ( \dfrac{1}{8} \right )^2 + 1 = \dfrac{1}{64} - \dfrac{18}{64} + \dfrac{64}{64} \ne 0.[/MATH]
[MATH]8 \left ( \dfrac{1}{4} \right )^3 - 18 \left ( \dfrac{1}{4} \right )^2 + 1 = \dfrac{1}{8} - \dfrac{18}{16} + \dfrac{8}{8} = \dfrac{1 + 8 - 9}{8} = 0.[/MATH]
[MATH]\therefore r = \dfrac{1}{4} \text { is one solution.}[/MATH]
[MATH]\dfrac{8r^3 - 18r^2 + 1}{r - \dfrac{1}{4}} = \dfrac{32r^3 - 72r^2 + 4}{4r - 1} = \dfrac{32r^3 - 8r^2 + 8r^2 - 72r^2 + 4}{4r - 1} =[/MATH]
[MATH]8r^2 - \dfrac{64r^2 - 4}{4r - 1} = 8r^2 - \dfrac{64r^2 - 16r + 16r - 4}{4r - 1} = 8r^2 - 16r - \dfrac{16r - 4}{4r - 1} = 8r^2 - 16r - 4.[/MATH]
[MATH]\left (r - \dfrac{1}{4} \right ) (8r^2 - 16r - 4) = 8r^3 - 16r^2 - 4r - 2r^2 + 4r + 1 = 8r^3 - 18r^2 + 1.[/MATH]
[MATH]16^2 - 4(8)(-\ 4) = 16(16 - 8) = 8 * 2 * 8(2 - 1) = 8^2 * 2 * 1 = 8^2 * 2.[/MATH]
[MATH]r = \dfrac{16 \pm \sqrt{8^2 * 2} }{2 * 8} = \dfrac{16 \pm 8\sqrt{2}}{16} = 1 \pm 0.5 \sqrt{2}.[/MATH]
At this point, you have three EXACT answers. The two involving the square root of 2 can be given a decimal approximation to any required degree of accuracy.

You do not need anything but algebra and arithmetic.

I'm not at all sure what you are doing here...I suppose you are going back to the original cubic? I used the Rational Roots Theorem to find candidate divisors and when I found a factor I used it to derive a quadratic and then used the quad formula to get the other two roots.

 

[JeffM]

I'm not at all sure what you are doing here...I suppose you are going back to the original cubic? I used the Rational Roots Theorem to find candidate divisors and when I found a factor I used it to derive a quadratic and then used the quad formula to get the other two roots.

I too am not quite sure what you are confused by.

The rational root theorem says that if you have a polynomial of degree n with integer coefficients and if the polynomial has a rational root, then the numerator of such a root is an integer factor of the constant term, and the denominator is an integer factor of the the coefficient of the high order term. (The latter explains the integer root theorem.) Therefore, for the given cubic, a rational root will have a numerator of 1 and a denominator of an integer factor of 8, of which four are positive and four negative.

So technically the possible rational roots OF THE CUBIC are

[MATH]\pm \dfrac{1}{1},\ \pm \dfrac{1}{2},\ \pm \dfrac{1}{4}, \text { and } \pm \dfrac{1}{8}.[/MATH]
However, plus and minus 1 are not roots by cursory inspection. Moreover, f(0) > 0, and f(1) < 0 so there must be at least one root between 0 and 1, meaning a positive root.

So I start by testing the positive fractions that are possible roots OF THE CUBIC.

1/8 does not work. 1/4 does. So we have a rational root. Now we use the fundamental theorem of algebra to find a polynomial of one fewer degree by division. In this case, we get a quadratic.

I then checked that I had divided correctly. . And then I used the quadratic formula to find the remaining roots.

So where are you confused?

 

[allegansveritatem]

I too am not quite sure what you are confused by.

The rational root theorem says that if you have a polynomial of degree n with integer coefficients and if the polynomial has a rational root, then the numerator of such a root is an integer factor of the constant term, and the denominator is an integer factor of the the coefficient of the high order term. (The latter explains the integer root theorem.) Therefore, for the given cubic, a rational root will have a numerator of 1 and a denominator of an integer factor of 8, of which four are positive and four negative.

So technically the possible rational roots OF THE CUBIC are

[MATH]\pm \dfrac{1}{1},\ \pm \dfrac{1}{2},\ \pm \dfrac{1}{4}, \text { and } \pm \dfrac{1}{8}.[/MATH]
However, plus and minus 1 are not roots by cursory inspection. Moreover, f(0) > 0, and f(1) < 0 so there must be at least one root between 0 and 1, meaning a positive root.

So I start by testing the positive fractions that are possible roots OF THE CUBIC.

1/8 does not work. 1/4 does. So we have a rational root. Now we use the fundamental theorem of algebra to find a polynomial of one fewer degree by division. In this case, we get a quadratic.

I then checked that I had divided correctly. . And then I used the quadratic formula to find the remaining roots.

So where are you confused?

well, now that you have explained it all so clearly, I can't say anymore what my problem was. Thanks