Question

How do you solve $ \sqrt[4]{{{x^4} + 1}} = 3x $ ?

Answer 1

Hint: In order to determine the solution of the given equation, simplify the equation by transposing the 4th root from LHS to RHS which will become $ {\left( {} \right)^4} $ in RHS. We will obtain a quartic equation. Combine like terms. Derive the value of variable $ x $ by solving the equation for $ x $ by transposing 4th power toward another side in form 4th root. Simplify the equation to obtain the required solution.

Complete step-by-step answer:
We are given an equation $ \sqrt[4]{{{x^4} + 1}} = 3x $ .
As we can see the term on the left-hand side contains the 4th root value. When transposing this 4th root towards the right-hand side it will become an exponent having value as $ {\left( {} \right)^4} $ . And similarly when we transpose the square root we have $ {\left( {} \right)^2} $ .
Now our equation becomes
 $ {x^4} + 1 = {\left( {3x} \right)^4} $
Simplifying the right-hand side term, we get
 $ {x^4} + 1 = 81{x^4} $
Combining like terms , we have
 $
  1 = 81{x^4} - {x^4} \\
  80{x^4} = 1 \;
  $
Dividing both sides of the equation by the coefficient of $ {x^4} $ i.e. 80
 $
  \dfrac{{80{x^4}}}{{80}} = \dfrac{1}{{80}} \\
  {x^4} = \dfrac{1}{{80}} \;
  $
Now transposing the 4th exponent from left-hand side to right-hand side, it will become the 4th root having both positive and negative values as $ \pm \sqrt[4]{{}} $ .
 $
  x = \pm \sqrt[4]{{\dfrac{1}{{80}}}} \\
  x = \pm \dfrac{1}{{\sqrt[4]{{80}}}} \;
  $
Now try to write the value 80 in the form of its prime factors, we get $ 80 = 2 \times 2 \times 2 \times 2 \times 5 $ . Replacing this in the equation
 $
  x = \pm \dfrac{1}{{\sqrt[4]{{2 \times 2 \times 2 \times 2 \times 5}}}} \\
  x = \pm \dfrac{1}{{2\sqrt[4]{5}}} \;
  $
Therefore, the required solution for the given equation is $ x = \pm \dfrac{1}{{2\sqrt[4]{5}}} $ .
So, the correct answer is “ $ x = \pm \dfrac{1}{{2\sqrt[4]{5}}} $ ”.

Note: 1. The value inside the root should be positive , if it is negative then it becomes an imaginary number.
2. The nth root of any number $ a $ is equivalent to $ \sqrt[n]{a} = {a^{\dfrac{1}{n}}} $ .
3. Don’t forget to cross-check your answer.
Quartic Equation: A Bi-Quartic equation is a equation which can be represented in the form of $ a{x^4} + b{x^3} + c{x^2} + dx + e = 0 $ where $ x $ is the unknown variable and a,b,c,d,e are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will be no more quartic.
The degree of the Bi-Quartic equation is of the order 4.
EveryBi-Quartic equation has 4 roots.