Question

There is a number x such that
$x {2}$


is irrational but
$x {4}$
is rational. Then, x can be what?​

Answer 1

First of all , the question you've written is wrong.

It must be

${x}^{2} \: \: and \: \: {x}^{4}$

but you've mentioned them as 2x and 4x

Answer:

there can be infinitely terms for x

Step-by-step explanation:

let us assume

$x = \sqrt[4]{6} \\ so \\ {x}^{2} = ( { \sqrt[4]{6} })^{2} \\ = \sqrt{6}$

√6 is irrational.

${x}^{4} = ( { \sqrt[4]{6} })^{4} \\ {x}^{4} = 6$

6 is rational.

hence, proved.

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