# Math Help - Algebra Question. Involving Exponents

1. ## Algebra Question. Involving Exponents

I Got this Question from my maths teacher.. Plz help.!!
X^1/3 + x^1/2 =12

I got the answer 64.. but by hit n trial, a meaningful one at that..but still show me how to solve it..PLZ!

TY!

2. Hello, umangarora!

$\text{Solve for }x\!:\;\;x^{\frac{1}{2}} + x^{\frac{1}{3}}\:=\:12$

Let $x^{\frac{1}{6}} \:=\:u \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:u^3 \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:u^2$

Then we have: . $u^3 + u^2 - 12 \:=\:0$

If this cubic has any rational roots,
. . they must be among: . $\pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12$

We find that $u = 2$ is a root.

Then: . $u^3 + u^2 - 12 \:=\:(u-2)(u^2 + 3u + 6)\:=\:0$
. . and the quadratic factor has no real roots.

So we have: . $u \,=\,2 \quad\Rightarrow\quad x^{\frac{1}{6}} \,=\,2 \quad\Rightarrow\quad x \:=\:64$

3. Originally Posted by umangarora
I Got this Question from my maths teacher.. Plz help.!!
X^1/3 + x^1/2 =12

I got the answer 64.. but by hit n trial, a meaningful one at that..but still show me how to solve it..PLZ!

TY!
A slightly simpler way (than working with rational powers) to solve is...

$\displaystyle\ x^{\frac{1}{3}}+x^{\frac{1}{2}}=12$

$\displaystyle\ y=x^{\frac{1}{3}}$

$\displaystyle\ y+y^{\frac{3}{2}}=12$

$\displaystyle\ y(1+\sqrt{y})=12$

$w=\sqrt{y}$

$w^2(1+w)=12$

$f(w)=w^3+w^2-12=0$

$f(2)=0$

$w=2\Rightarrow\ y=w^2=4\Rightarrow\ x=y^3=64$

4. exactly Guys..
what u have done is factored.. but u did use hit and trial somewhere.. is there NO proof which boils down to x=64..?
Do u have to put 2 as u or x^1/3
to get the answer :S ?
P.s. i did the same thing.
what i did is:
x^1/2 + x^1/3 =12
since rhs is rational lhs has to be rational.
there for x has to be m^3 and n^ 2
min. value is when x=m^6
trying 1, we get 2.
trying 2 we get the answer.
trying 3, we exceed.

But since the Q includes cube root, we must try negative value,
trying -1, we get 0
trying -2, we get -4
trying -3, we exceed.
so 2 is the only real ans.

5. Originally Posted by umangarora
exactly Guys..
what u have done is factored.. but u did use hit and trial somewhere.. is there NO proof which boils down to x=64..?
Do u have to put 2 as u or x^1/3
to get the answer :S ?
P.s. i did the same thing.
what i did is:
x^1/2 + x^1/3 =12
since rhs is rational lhs has to be rational.
there for x has to be m^3 and n^ 2
min. value is when x=m^6
trying 1, we get 2.
trying 2 we get the answer.
trying 3, we exceed.

But since the Q includes cube root, we must try negative value,
trying -1, we get 0
trying -2, we get -4
trying -3, we exceed.
so 2 is the only real ans.
Apart from the techniques to solve cubics,

using a little logic, we can say for $w^3+w^2=12$

$|w|>1$

therefore if $w$ is negative, then $w(w^2+w)$ is negative.
Negative solutions do not apply.

Then, if you look at $w(w^2+w)$

you see that if $w<2,$ the product is $<12$

and if $w>2,$ the product is $>12$