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