Question

(\sqrt[6]{x} )^2-3\sqrt[6]{x} =4

Answer 1

$(\sqrt[6]{x} )^2-3\sqrt[6]{x} =4\\\\\sqrt[6]{x}=m,m\geq 0\\\\m^{2}-3m-4=0\\\\D=(-3)^{2}-4*(-4)=9+16=25=5^{2}\\\\m_{1}=\frac{3+5}{2}=4\\\\m_{2}=\frac{3-5}{2}=-1$

корень m₂ - не подходит, так как < 0

$\sqrt[6]{x}=4\\\\(\sqrt[6]{x})^{6}=4^{6}\\\\x=4096$

Answer 2

Let's solve the equation step by step. The equation is:

$(\sqrt[6]{x})^2 - 3\sqrt[6]{x} = 4$

To simplify, let's make a substitution: let $y = \sqrt[6]{x}$. Now the equation becomes:

$y^2 - 3y = 4$

To solve this quadratic equation, we need to bring all terms to one side of the equation, set it equal to zero, and then factor it (if possible) or use the quadratic formula.

$y^2 - 3y - 4 = 0$

Now, let's factor this quadratic equation:

$(y - 4)(y + 1) = 0$

Now we set each factor equal to zero and solve for $y$:

1. $y - 4 = 0$ ---> $y = 4$

2. $y + 1 = 0$ ---> $y = -1$

Remember, $y = \sqrt[6]{x}$, so we have two possible values for $y$:

1. $y = 4$ ---> $\sqrt[6]{x} = 4$

To get the value of $x$, raise both sides to the power of 6:

$x = 4^6 = 4096$

2. $y = -1$ ---> $\sqrt[6]{x} = -1$

However, the sixth root of a number can't be negative since any real number raised to a positive power will yield a positive result. Therefore, there is no solution for $y = -1$ in the real domain, which means there is no solution for $x$ in this case.

So, the only solution to the original equation is $x = 4096$.