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,

$-2x=-0.875$

$x=0.4375$

$⇒a=0.5625$