An equation with negative exponents in quadratic equations test

Question

There is a problem like this :

$x^{-1} = 2x^{(-1/2)} + 3 , x = $?

in my test. I'm working on it for a half of hour but still i can't solve. Please help me. (Excuse my bad grammar)

Answer 1

since $$ x^{-1} = 2x^{-\frac12} +3 $$ by squaring: $$ x^{-2} = 4x^{-1} +12x^{-\frac12} +9 \\ = 4x^{-1} +12\frac{x^{-1}-3}2 +9\\ =10x^{-1} -9 $$ i.e. $$ x^{-2} -10x^{-1}+9 =0 $$ or $$ (x^{-1}-9)(x^{-1}-1) = 0 $$

Answer 2

If you define $y=\frac1{\sqrt{x}}$, then you can find $y$ from the equation, since

$$x^{-1} = 2x^{-\frac12} + 3$$

implies $$y^2 = 2y + 3$$

and you can find the values of $y$ quite quickly.

Every value of $y$ then gives you a candidate value for x which you must check for correctness by plugging it into the original equation.