Question

# Solve the following equations: $3{x^{\dfrac{1}{{2n}}}} - {x^{\dfrac{1}{n}}} - 2 = 0$.

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Hint: Let’s try to convert the given equation in the form of an equation which can be solved (i.e, in this case as a quadratic equation) by doing some substitution and afterwards, solve for these substitution we’ll get the answer.

Given, $3{x^{\dfrac{1}{{2n}}}} - {x^{\dfrac{1}{n}}} - 2 = 0$

If we put ${x^{\dfrac{1}{{2n}}}} = t$, the above equation reduces to a quadratic equation.

${x^{\dfrac{1}{{2n}}}} = t...................{\text{(1)}}$

Squaring both sides of the above equation, we get,

${\left( {{x^{\dfrac{1}{{2n}}}}} \right)^2} = {t^2} \Rightarrow {x^{\dfrac{1}{n}}} = {t^2}....................(2)$

Substituting equations (1) and (2) in the given equation, the quadratic equation obtained is,

$3t - {t^2} - 2 = 0$

Multiplying the above equation by - 1 on both sides, we get,

${t^2} - 3t + 2 = 0$

Now, solving the above quadratic equation by factorization method,

We can write, ${t^2} - t - 2t + 2 = 0$

Taking 't' common from first two terms and ' - 2' common from last two terms on LHS of the above equation,

$\Rightarrow t\left( {t - 1} \right) - 2\left( {t - 1} \right) = 0$

Now, taking $\left( {t - 1} \right)$ common from both the terms on LHS of the above equation,

$\Rightarrow \left( {t - 1} \right)\left( {t - 2} \right) = 0$

Either $t - 1 = 0$ or $t - 2 = 0$

$\Rightarrow$ t = 1 or t = 2

According to equation (1), we have

${x^{\dfrac{1}{{2n}}}} = t$

Either t = 1 then ${x^{\dfrac{1}{{2n}}}} = 1$ or t = 2 then ${x^{\dfrac{1}{{2n}}}} = 2$

$\Rightarrow$ $(x^{\dfrac{1}{2n}})^{2n} = 1^{2n}$ or

$\Rightarrow$ $(x^{\dfrac{1}{2n}})^{2n} = 2^{2n}$

Since, ${\left( 1 \right)^{2n}} = 1{\text{ and }}{\left( 2 \right)^{2n}} = {\left( {{2^2}} \right)^n}$

$\Rightarrow x = 1{\text{ or }}x = {\left( 4 \right)^n}$

Hence, the solution of the given equation is either x = 1 or x = ${\left( 4 \right)^n}.$

Note: - Since we are dealing with fractions in exponents, be careful with calculations.