Question 948047
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And yet another hasty, careless, and incorrect response from tutor @CubeyThePenguin.<br>
You can introduce a dummy variable, as he does in his response, to help solve the problem if you want; it is not necessary.<br>
{{{x^(2/5)-3x^(1/5)-4 = 0}}}
{{{(x^(1/5)-4)(x^(1/5)+1) = 0}}}
{{{x^(1/5)=4}}} or {{{x^(1/5) = -1}}}
{{{x = 4^5 = 1024}}} or {{{x = (-1)^5 = -1}}}<br>
The x = -1 solution is NOT extraneous; it satisfies the original equation.<br>