1. ## Fractional indices

Help needed with the following question

$x^{\frac{1}{3}} + 9x^{-\frac{1}{3}} = 4$

Solve for x .

2. Multiply through by $x^{1/3}$

$(x^{1/3})^2 + 9 = 4x^{1/3}$ and rearrange to give $\displaystyle \left(x^{1/3}\right)^2 - 4x^{1/3} + 9 = 0$ which is a quadratic equation.

I got

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

Then,
y^{2}-4y+9=0$

I dont know what to do after that..... I used the quadratic formula...but that didnt work....it's giving imaginary values but the answer consists of real values

4. Originally Posted by Arka

I got

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

Then,
y^{2}-4y+9=0$

I dont know what to do after that..... I used the quadratic formula...but that didnt work....it's giving imaginary values but the answer consists of real values
Everything has been done correctly - you must have typed the question wrong.

5. Originally Posted by Arka
$Let y=x^{\frac{1}{3}}
Yes, that is correct- completing the square, $y^2- 4y+ 4- 4+ 9= 0$ becomes $(y- 2)^2= -5$. That has NO real roots which means that the original equation, $x^{1/3}- 9x^{-1/3}= 4$has no real roots. Either you have copied the equation incorrectly or the answer is "there is no real number solution".