Algebra Question

The expression is 16/a - 8a + a^3 = b^2

To solve for A you need to have another equation with a and b in to get a numerical answer.

Getting an algebraic answer for a in terms of b is possible but 16(a^-1) - 8a +(a^3) = b^2 is not easily do-able off the top of my head and I think you'd need to compute a numerical solution for a in terms of b which isn't ideal and isn't really algebra.

What level is this question meant for? If it's high school then the question is wrong/you've not given us all the information, if it's college level or higher then it's potentially the right question but not really an algebra question.

no, when there are variables of diffrent powers there is no way (as far as algebra level math students are concerned) to reduce to just A.

there are cube root equations that are a mile long...

find a book on classical algebra and punch in the numbers..

This is a question for a college level course

How about this form of the expression then: (4 - a^2) / a^1/2 = b

Multiply through by a.

The powers of a are now a^4 and a^2.

Let x=a^2.

You get a quadratic equation in x.

Remember that b is a constant.

Use the quadratic formula to get the two solutions for x.

Don't forget +- in quadratic formula.

(in terms of b).

Take the square root of these solutions,

don't forget +-.

So you should have 4 solutions in all.

Thales wrote:

Multiply through by a.

The powers of a are now a^4 and a^2.quote]

Wrong, the equation is now 16-8a^2+ a^4 = a(b^2)

[quote]Thales wrote:

Let x=a^2.

You get a quadratic equation in x.

No, you get 16- 8x + x^2 = (x^.5)(b^2)

Still not any closer to a solution. Good try though

Three Squaws

Three squaws were each preparing for the birth of their first child. The

first squaw placed a large bear hide by a river, the second squaw placed

an elk hide by a tree by a river, and the third squaw placed a

hippopotamus hide by a path, near the river and the tree so that the

three formed a triangle.

It just so happens that all three women gave birth on the same day.

The first squaw on the bear hide had a 5-lb son, the second squaw on the

elk hide had a 6-lb son, and the third squaw on the hippopotamus hide had

an 11-lb son.

To this day, mathematicians credit these three women with the first proof of the Pythagorean Theorem:

"The son of the squaw of the hippopotamus is equal to the sons

of the squaws of the two adjacent hides."

ner wrote:

Thales wrote:

Multiply through by a.

The powers of a are now a^4 and a^2.quote]

Wrong, the equation is now 16-8a^2+ a^4 = a(b^2)

[quote]Thales wrote:

Let x=a^2.

You get a quadratic equation in x.

No, you get 16- 8x + x^2 = (x^.5)(b^2)

Still not any closer to a solution. Good try though

Sure you are. Then you've got:

(x-4)^2 = x^(1/2)b^2

(x-4)^2x^(-1/2) = b^2

( x^(3/4) - 4 x^(-1/4) ) ^2 = b^2

then unsquare both sides (watch out for +/-) and solve from there.

With the information you gave us you can solve for a by imagining b is anything. To solve for both, you will need another equation. Your question is incomplete.

He said in the OP that he only needed to solve for A.

6.

The following comes from my fourteen year old son:

I see your problem solving this. Without at least one more equation with the variables "a" and "b" it is impossible to solve (with another equation, you would simply solve for "b" in terms of "a" and substitute into the original equation). However, seeing as you DON'T have a second equation, the best you can possibly hope for is to solve for "a" in terms of "b." When you do that, you come up with the answer:

"b=9/16a^-1", or, to put it more more simply, "-9/16a=b."

If you are going to show your work for this equation, GOOD LUCK; it involves about 10 steps.

montesquieu,

your son is wrong. let a=1, then b=-9/16. plug that in and you get 16/1 - 8*1 + 1^3 = 81/256. or, 9 = 81/256. clearly not true.

basically, you have a quartic equation. according to wikipedia, you can solve it by radicals. i would solve it by using the computer software someone else wrote to save me the headache. i plugged it into mathematica and here is the answer:

\!\(\(\(x -> \(-\(1\/2\)\)\

√\((16\/3 + \(256\

2\^\(1/3\)\)\/\(3\ \((8192 + 27\

b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) + \

\((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\

\^\(1/3\)\))\) - 1\/2\ √\((32\/3 - \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ \

b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) - \((8192 + 27\ \

b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\) - \

\((2\ b\^2)\)/\((√\((

16\/3 + \(256\ 2\^\(1/

3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ \

b\^8\))\)\^\(1/3\)\) + \((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ \

b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\))\))\))\)\)\(}\)\), {x -> \(-\(1\/2\)\)\

√\((16\/3 + \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \

\@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) + \((8192 + 27\ b\^4 + 3\ \@3\

\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\))\) + 1\/2\ √\((

32\/3 - \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \

\@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) - \((8192 + 27\ b\^4 + 3\ \@3\

\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\) - \((2\ \

b\^2)\)/\((√\((16\/3 + \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \@3\ \

\@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/

3\)\) + \((

8192 + 27\ b\^4 +

3\ \@3\ \@\(16384\ b\^4 + 27\ \

b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\))\))\))\)}, {x -> 1\/2\ √\((16\/3 + \

\(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ \

b\^8\))\)\^\(1/

3\)\) + \((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ \

b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\))\) - 1\/2\ √\((32\/3 - \(256\

\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\

b\^8\))\)\^\(1/3\)\) - \((8192 + 27\

b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\

\^\(1/3\)\) + \((2\ b\^2)\)/\((√\((16\/3 + \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 \

+ 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) + \((8192 + \

27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ \

2\^\(1/3\)\))\))\))\)}, {x -> 1\/2\ √\((16\/3 + \(256\ 2\^\(1/3\)\)\/\(3\ \

\((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) + \

\((8192 + 27\ b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\

\^\(1/3\)\))\) + 1\/2\ √\((32\/3 - \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ \

b\^4 + 3\ \@3\ \@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) - \((8192 + 27\ \

b\^4 + 3\ \@3\ \@\(16384\ b\^4 +

27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\) + \((2\ \

b\^2)\)/\((√\((16\/3 + \(256\ 2\^\(1/3\)\)\/\(3\ \((8192 + 27\ b\^4 + 3\ \@3\ \

\@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\) + \((8192 + 27\ b\^4 + 3\ \@3\ \

\@\(16384\ b\^4 + 27\ b\^8\))\)\^\(1/3\)\/\(3\ 2\^\(1/3\)\)\)\)\)\)

obviously that looks screwy because the board is not set up for copying and pasting from mathematica. but solving that thing by hand is not a reasonable task

Montesquieu wrote:

The following comes from my fourteen year old son:

I see your problem solving this. Without at least one more equation with the variables "a" and "b" it is impossible to solve (with another equation, you would simply solve for "b" in terms of "a" and substitute into the original equation). However, seeing as you DON'T have a second equation, the best you can possibly hope for is to solve for "a" in terms of "b." When you do that, you come up with the answer:

"b=9/16a^-1", or, to put it more more simply, "-9/16a=b."

If you are going to show your work for this equation, GOOD LUCK; it involves about 10 steps.

I bet you think you're real clever posting that your 14 year old son could do this while I'm sure I'm not the only one with a graduate degree in math to consider the problem.

Basically, the solution will almost surely involve complex roots, as I reduced it to a quartic and plugged in some sample values of b into a quartic root calculator. It takes powerful software like mathematica to evaluate analytically.

Stop trolling.

the "@" are square roots. i don't know why some copied correctly and other didn't.

I messed up. Sorry.

You've got to use the quartic formula, which depends on the cubic formula.

We can write the eqn as a^2-4=b*sqrt(a). Let x=sqrt(a).

You get the quartic

x^4-bx-4=0.

Here is the solution. (See mathworld.com for more explanation..)

First solve the equation

(u/4)+(4/u)=(b^2/u^2).

If b=0 this is easy. Otherwise, multiply out and get the

"resolvent cubic": u^3+16u=4b^2.

This is solved as follows;

Let w be a cube root of

2b^2+sqrt(4b^4+(16^3/27)).

Then u=w-(16/3w).

Now let v=sqrt(u).

With this v in hand, the solution x is a root of the quadratic equation

x^2-vx+(v^2/2)-(b/v)=0.

Finally, a=x^2.

Not too enlightening, but there it is.

Brilliant!