Question

Solve the equation ${{x}^{\dfrac{2}{n}}}+6=5{{x}^{\dfrac{1}{n}}}$.

Answer 1

Hint: Substitute $t={{x}^{\dfrac{1}{n}}}$ and ${{t}^{2}}={{x}^{\dfrac{2}{n}}}$.This will give us a quadratic equation in $t$. Solve this quadratic equation and find the values of $t$. Then using these values of $t$ and $t={{x}^{\dfrac{1}{n}}}$ , find the required values of $x$.

Complete step-by-step answer:

In this question, we need to solve the equation ${{x}^{\dfrac{2}{n}}}+6=5{{x}^{\dfrac{1}{n}}}$ for $x$.
In this question, it can be a bit confusing to carry on with the terms ${{x}^{\dfrac{2}{n}}}$ and $5{{x}^{\dfrac{1}{n}}}$.
So, to make it easier we will substitute $t={{x}^{\dfrac{1}{n}}}$. Then find the values of $t$ which will further give us the values of $x$.
This gives us ${{t}^{2}}={{x}^{\dfrac{2}{n}}}$ .
Substituting these in the given equation , we get the following:
${{t}^{2}}+6=5t$
Rearranging the terms, we will get the following:
${{t}^{2}}-5t+6=0$
Now, we have a quadratic equation in $t$. We need to find the roots of this quadratic equation to find the answer.
To find the roots, we will expand the middle term.
We can write $5t$ as $2t+3t$.
Substituting this in the above equation, we get the following:
${{t}^{2}}-2t-3t+6=0$
Now, we will take $t$common from the first two terms and we will take 3 common from the last two terms.
After doing this, we will get the following:
$t\left( t-2 \right)-3\left( t-2 \right)=0$
Now we will take $\left( t-2 \right)$ common from both these terms.
After doing this, we will get the following:
$\left( t-2 \right)\left( t-3 \right)=0$
From this, we get the roots of the equation in t:
$t=2,3$
Now, we substituted $t={{x}^{\dfrac{1}{n}}}$ before.
Using this, we will find the values of $x$.
Substituting $t=2,3$ in $t={{x}^{\dfrac{1}{n}}}$ , we will get the following
$2={{x}^{\dfrac{1}{n}}}$ and $3={{x}^{\dfrac{1}{n}}}$
Raising this to the power of n, we get the following:
$x={{2}^{n}}$ and $x={{3}^{n}}$
This is our final answer.

Note: In this question, it can be a bit confusing to carry on with the terms ${{x}^{\dfrac{2}{n}}}$ and $5{{x}^{\dfrac{1}{n}}}$. So, to make it easier we have substituted $t={{x}^{\dfrac{1}{n}}}$ . This will give us a quadratic equation in $t$ which can be solved easily. Solve this equation and then find the values of $t$which will further give us the values of $x$.