Baffled by old school exam

[BvU]

Homework Statement

Solve $x$ from $$\left ( a+x\right )^{2\over 3} + 6\left ( a-x\right )^{2\over 3} =5\left ( a^2- x^2 \right )^{1\over 3} $$

Relevant Equations

actually, my algebra appears insufficient :frown:

Book answer is $\qquad a≥0\qquad x={ 7\over 9}\;a\ \ \lor\ \ x={ 13\over 14}\;a\ \$but I fail to see how to get there !

Stunned by an 1886 dutch high school exam exercise. Hats off for the 17 year olds that did it !

$\$

 

[Frabjous]

Divide by the right hand side.

 

[phinds]

Divide by the right hand side.

I don't get it. The right side does not divide at all nicely into the left side. Did you overlook the exponents inside the parens or am I missing something?

 

[Steve4Physics]

Homework Statement: Solve $x$ from $$\left ( a+x\right )^{2\over 3} + 6\left ( a-x\right )^{2\over 3} =5\left ( a^2- x^2 \right )^{1\over 3} $$
Relevant Equations: actually, my algebra appears insufficient

Book answer is $\qquad a≥0\qquad x={ 7\over 9}\;a\ \ \lor\ \ x={ 13\over 14}\;a\ \$but I fail to see how to get there !

Stunned by an 1886 dutch high school exam exercise. Hats off for the 17 year olds that did it !

$\$

I think this should work:

Let $p = (a+x)^{\frac 13}$ and $q = (a-x)^{\frac 13}$.

The given equation can then be written as $p^2 + 6q^2 =5pq$.

$p^2 - 5pq + 6q^2 =0$

$(p - 2q)(p - 3q) = 0$

And take it from there.

Edit: Spoiler removed.

 

[BvU]

Divide by the right hand side.

I should have seen it ... but didn't : the quadratic equation is mischieviously wrapped up in that kind of exam question.

@phinds : divide by $(a-x)^{1\over 3} \ (a+x)^{1\over 3}$ to get $y+{6\over y} =5$

$\$

 

[hutchphd]

As soon as I appreciated that $$(a+x)(a-x)=(a^2-x^2)$$ I was home free on this one

 

[WWGD]

Worse comes to worse, time is running out on the exam, cube both sides. Only one term will remain to a non-unity power p/q. Leave it alone in one side and raise both sides to the qth power. Hardly elegant, but at least you'll get some credit.
Edit: In this exercise, you first cube both sides, end up with a mixed term; then isolate it on one side, then cube both sides again.

 

[BvU]

Worse comes to worse, time is running out ...

It's an exam from 1886

 

[SammyS]

It's an exam from 1886

It's likely too late for you to get any credit for it now.

 

[WWGD]

It's an exam from 1886

But it was an _open_ exam. ;).

 

[gmax137]

Worse comes to worse, time is running out

 

[vanhees71]

Homework Statement: Solve $x$ from $$\left ( a+x\right )^{2\over 3} + 6\left ( a-x\right )^{2\over 3} =5\left ( a^2- x^2 \right )^{1\over 3} $$
Relevant Equations: actually, my algebra appears insufficient

Book answer is $\qquad a≥0\qquad x={ 7\over 9}\;a\ \ \lor\ \ x={ 13\over 14}\;a\ \$but I fail to see how to get there !

Stunned by an 1886 dutch high school exam exercise. Hats off for the 17 year olds that did it !

$\$

There you see how much the level of high-school math has declined nowadays! SCNR.

 

[DrDu]

There you see how much the level of high-school math has declined nowadays! SCNR.

No wonder:
https://www.sciencedaily.com/releases/2022/03/220307162011.htm

 

[vanhees71]

I don't know, whether it's about IQ. It's rather about quality of high-school education. A colleague just told me in his last problem session a 4th semester (sic!) physics student was utmost lost at the task to do an integral by partial integration.

 

[lurflurf]

I don't know, whether it's about IQ. It's rather about quality of high-school education. A colleague just told me in his last problem session a 4th semester (sic!) physics student was utmost lost at the task to do an integral by partial integration.

What is partial integration? You mean like this? $$\int \mathrm{f}(x,y,z) \partial{}x$$
or this
$$\int u\, dv=u\, v-\int v\, du$$
maybe your friend should work on his communication skills. The student probably did not know what he was on about. What is forth semester physics these days? Is that still incline planes or maybe finite square wells?

 

[vanhees71]

The first expression is nonsensical. Of course I mean the second.

 

[Steve4Physics]

The first expression is nonsensical. Of course I mean the second.

It may be worth noting that the term ‘partial integration’ has two (that I know of) different meanings:

1) It is an alternative (though not widely used) name for ‘integration by parts’.

2) It is a term (informally?) used in multivariate calculus for integration with respect to a single variable. For example in https://www.cliffsnotes.com/study-guides/differential-equations/review-and-introduction/partial-integration the author writes $f(x,y) = \int M(x,y)∂x$ (which is consistent with @lurflurf's 1st example in Post #15). Though I don't know if this notational use of $∂x$ is widely accepted.

 

[chwala]

What is partial integration? You mean like this? $$\int \mathrm{f}(x,y,z) \partial{}x$$
or this
$$\int u\, dv=u\, v-\int v\, du$$
maybe your friend should work on his communication skills. The student probably did not know what he was on about. What is forth semester physics these days? Is that still incline planes or maybe finite square wells?

I thought partial integration and integration by parts mean two different things...correct me if i am wrong...i think partial implies to for e.g decomposing

$\dfrac{ax+c}{ax^2+bx+c}$ into $\dfrac{A}{(ax+b)} +\dfrac{B}{(cx+d)}$ where $x$ is a given variable and the rest are constants.

...then applying integration accordingly.

 

[chwala]

Homework Statement: Solve $x$ from $$\left ( a+x\right )^{2\over 3} + 6\left ( a-x\right )^{2\over 3} =5\left ( a^2- x^2 \right )^{1\over 3} $$
Relevant Equations: actually, my algebra appears insufficient

Book answer is $\qquad a≥0\qquad x={ 7\over 9}\;a\ \ \lor\ \ x={ 13\over 14}\;a\ \$but I fail to see how to get there !

Stunned by an 1886 dutch high school exam exercise. Hats off for the 17 year olds that did it !

$\$

It would be interesting if i can get a different approach to this problem...i will check it out. I guess Binomial Theorem could apply though tedious.

Ok i am thinking along the lines of:

$\dfrac{(a+x)^\frac{2}{3}}{(a+x)^\frac{1}{3}}= 5(a-x)^\frac{1}{3} - 6(a-x)^\frac{2}{3}$$(a+x)^\frac{1}{3}=5(a-x)^\frac{1}{3} - 6(a-x)^\frac{2}{3}$$\dfrac {(a-x)^\frac{1}{3}[5-6(a-x)^\frac{1}{3}] }{(a+x)^\frac{1}{3}}=1$

Let

$a-x=k$
$a+x=m$

Then using the above... we shall get:

$6k^\frac{2}{3}-5k^\frac{1}{3}+1=0$

and

$m^\frac{1}{3}=1$

Let

$n=k^\frac{1}{3}$

then we shall have:

$6n^2-5n+1=0$

$n_1=0.5 ⇒k_1=0.125$

$n_2=0.333⇒k_2=0.037$

We know that

$m=1$

using, $k=0.125$ we shall have,

$a-x=0.125$
$a+x=1$

solving the simultaneous equation yields,

$-2x=-0.875$

$x=0.4375$

$⇒a=0.5625$

 

[brotherbobby]

I hope I am not too late, but I have found the answers you put for $x$ but not sure why $a$ should be greater or less than zero, as you say in your first post (#1 up).

I hope this is readable.

Can anyone tell me please why should $a\ge 0$? And if am mistaken in my work up there ?

 

[BvU]

What do you get if $a = -1$ ?
check

And high school curriculum didn't include complex numbers

$\$

 

[SammyS]

It would be interesting if i can get a different approach to this problem...i will check it out. I guess Binomial Theorem could apply though tedious.

Ok i am thinking along the lines of:

$\dfrac{(a+x)^\frac{2}{3}}{(a+x)^\frac{1}{3}}= 5(a-x)^\frac{1}{3} - 6(a-x)^\frac{2}{3}$

It looks like you intended to divide both sides buy $(a+x)^\frac{1}{3}$ but you neglected dividing $5(a-x)^\frac{1}{3}$ by $(a+x)^\frac{1}{3}$ .

 

[chwala]

It looks like you intended to divide both sides buy $(a+x)^\frac{1}{3}$ but you neglected dividing $5(a-x)^\frac{1}{3}$ by $(a+x)^\frac{1}{3}$ .

...I will check this again...

 

[chwala]

This

It would be interesting if i can get a different approach to this problem...i will check it out. I guess Binomial Theorem could apply though tedious.

Ok i am thinking along the lines of:

$\dfrac{(a+x)^\frac{2}{3}}{(a+x)^\frac{1}{3}}= 5(a-x)^\frac{1}{3} - 6(a-x)^\frac{2}{3}$$(a+x)^\frac{1}{3}=5(a-x)^\frac{1}{3} - 6(a-x)^\frac{2}{3}$$\dfrac {(a-x)^\frac{1}{3}[5-6(a-x)^\frac{1}{3}] }{(a+x)^\frac{1}{3}}=1$

Let

$a-x=k$
$a+x=m$

Then using the above... we shall get:

$6k^\frac{2}{3}-5k^\frac{1}{3}+1=0$

and

$m^\frac{1}{3}=1$

Let

$n=k^\frac{1}{3}$

then we shall have:

$6n^2-5n+1=0$

$n_1=0.5 ⇒k_1=0.125$

$n_2=0.333⇒k_2=0.037$

We know that

$m=1$

using, $k=0.125$ we shall have,

$a-x=0.125$
$a+x=1$

solving the simultaneous equation yields,

$-2x=-0.875$

$x=0.4375$

$⇒a=0.5625$

This may be unorthodox approach but hey let us do it!

We have,

$(a+x)^{\frac{2}{3}} + 6 (a-x)^{\frac{2}{3}}=5(a^2-x^2)^{\frac{1}{3}}$

I will still let

$m=a+x$

and

$k=a-x$

then,

$m^{\frac{2}{3}} + 6k^{\frac{2}{3}}=5m^{\frac{1}{3}}k^{\frac{1}{3}}$

$6k^{\frac{2}{3}}=m^{\frac{1}{3}}[5k^{\frac{1}{3}}-m^{\frac{1}{3}}]$

Let

$m^{\frac{1}{3}}=1$

then,

$6k^{\frac{2}{3}}-5k^{\frac{1}{3}}+1=0$

...

the other steps to solution shall remain as shown on the quoted post.

cheers!

 

[Frabjous]

ThisThis may be unorthodox approach but hey let us do it!

We have,

$(a+x)^{\frac{2}{3}} + 6 (a-x)^{\frac{2}{3}}=5(a^2-x^2)^{\frac{1}{3}}$

I will still let

$m=a+x$

and

$k=a-x$

then,

$m^{\frac{2}{3}} + 6k^{\frac{2}{3}}=5m^{\frac{1}{3}}k^{\frac{1}{3}}$

$6k^{\frac{2}{3}}=m^{\frac{1}{3}}[5k^{\frac{1}{3}}-m^{\frac{1}{3}}]$

Let

$m^{\frac{1}{3}}=1$

then,

$6k^{\frac{2}{3}}-5k^{\frac{1}{3}}+1=0$

...

the other steps to solution shall remain as shown on the quoted post.

cheers!

Why define m^1/3=1?

 

[chwala]

Why define m^1/3=1?

If you look at the solution, one variable, that is $x$ is dependant on the other variable $a$...I can therefore choose to let my variable be equal to a certain value. The intention being to find any possible means to solve the equation (in my case using quadratic method).

Or is my thinking not correct?

 

[Mark44]

What is partial integration?

Based on @vanhees71's subsequent response, he evidently meant "integration by parts." It's possible that he conflated the names of the techniques of integration by parts and integration by partial fraction decomposition.

 

[Frabjous]

If you look at the solution, one variable, that is $x$ is dependant on the other variable $a$...I can therefore choose to let my variable be equal to a certain value. The intention being to find any possible means to solve the equation (in my case using quadratic method).

Or is my thinking not correct?

a is a constant and you were asked to solve for x(a).

You’ve added an equation, so you are solving for an intersection.

 

[chwala]

a is a constant and you were asked to solve for x(a).

You’ve added an equation, so you are solving for an intersection.

@Frabjous that is true, I added an equation which led me to the final solution in terms of $a$ and $x$. Check my post $19$ that has the continuation after i came up with the quadratic equation.

 

[Frabjous]

@Frabjous that is true, I added an equation which led me to the final solution. Check my post $19$ that has the continuation after i came up with the quadratic equation.

So what? You get one value of x for one value of a. The OP has the solution which has TWO values of x for a given value of a. The OP also has the solution for other values of a. You are missing a lot of solutions.

 

[chwala]

So what? You get one value of x for one value of a. The OP has the solution which has TWO values of x for a given value of a. The OP also has the solution for other values of a. You are missing a lot of solutions.

I will check if I can get other values by changing my $m$ value. Cheers man!

 

[Frabjous]

You are going to come up with a finite number number ‘n‘ of solutions. Given that there are an infinite number of solutions, your grade for the problem will be n/∞=0.

 

[SammyS]

@chwala ,
Your response to my post (#22), although I made a mistake in that post of mine. It is corrected here:

(My post (#22) in reference to your post (#19) )
It looks like you intended to divide both sides buy $(a+x)^\frac{1}{3}$ but you neglected dividing $5(a-x)^\frac{1}{3}$ by $(a+x)^\frac{1}{3}$ .

Correction:
... you neglected to divide $6(a-x)^\frac{2}{3}$ by $(a+x)^\frac{1}{3}$ .

Your post(#23)

This may be unorthodox approach but hey let us do it!

We have,

$(a+x)^{\frac{2}{3}} + 6 (a-x)^{\frac{2}{3}}=5(a^2-x^2)^{\frac{1}{3}}$

I will still let

$m=a+x$

and

$k=a-x$

then,

$m^{\frac{2}{3}} + 6k^{\frac{2}{3}}=5m^{\frac{1}{3}}k^{\frac{1}{3}}$

(I'll cut in here.)

This looks fine. In fact you could write that as;

$\displaystyle m^{\frac{2}{3}} -5m^{\frac{1}{3}}k^{\frac{1}{3}}+ 6k^{\frac{2}{3}}=0$ ,

which can be factored quite nicely, by the way.

You then go on to state;

$6k^{\frac{2}{3}}=m^{\frac{1}{3}}[5k^{\frac{1}{3}}-m^{\frac{1}{3}}]$

Let

$m^{\frac{1}{3}}=1$

then,

$6k^{\frac{2}{3}}-5k^{\frac{1}{3}}+1=0$
...

the other steps to solution shall remain as shown on the quoted post.

cheers!

Although you get the same quadratic equation as you did in the "quoted post" (#19), that is only due to the choice of $m=1$.

For other values of $m$, posts 19 and 22 will not agree.

 

[chwala]

@chwala ,
Your response to my post (#22), although I made a mistake in that post of mine. It is corrected here:Your post(#23)
(I'll cut in here.)

This looks fine. In fact you could write that as;

$\displaystyle m^{\frac{2}{3}} -5m^{\frac{1}{3}}k^{\frac{1}{3}}+ 6k^{\frac{2}{3}}=0$ ,

which can be factored quite nicely, by the way.

You then go on to state;

Although you get the same quadratic equation as you did in the "quoted post" (#19), that is only due to the choice of $m=1$.

For other values of $m$, posts 19 and 22 will not agree.

Thanks can see that we end up with what was previously found by other members:

... i.e

$x^2-5xy+6y^2=0$

$(x-2y)(x-3y)=0$

...

where;

$x=m^\frac{1}{3}$

and

$y=k^\frac{1}{3}$

 

[vanhees71]

Based on @vanhees71's subsequent response, he evidently meant "integration by parts." It's possible that he conflated the names of the techniques of integration by parts and integration by partial fraction decomposition.

I think it's a "Germanism", because in German we call this technique "partielle Integration". Obviously in English it's exclusively called "integration by parts".