Solve the equation $$\sqrt{x} - 3\cdot\sqrt[4]{x} - 10 = 0$$

Let's solve the equation $$\sqrt{x} - 3\sqrt[4]{x} - 10 = 0$$. First, let's make a substitution to simplify the equation. Let $$y = \sqrt[4]{x}$$. Then, $$y^2 = (\sqrt[4]{x})^2 = \sqrt{x}$$. Substituting these into the original equation, we get: $$y^2 - 3y - 10 = 0$$ Now we have a quadratic equation in terms of $$y$$. We can solve this by factoring. We are looking for two numbers that multiply to -10 and add up to -3. Those numbers are -5 and 2. So, we can factor the quadratic equation as: $$(y - 5)(y + 2) = 0$$ This gives us two possible solutions for $$y$$: $$y - 5 = 0 \Rightarrow y = 5$$ $$y + 2 = 0 \Rightarrow y = -2$$ Now, we need to substitute back to find the values of $$x$$. Remember that $$y = \sqrt[4]{x}$$. For $$y = 5$$: $$\sqrt[4]{x} = 5$$ Raise both sides to the power of 4: $$(\sqrt[4]{x})^4 = 5^4$$ $$x = 625$$ For $$y = -2$$: $$\sqrt[4]{x} = -2$$ Since the fourth root of a real number cannot be negative, there is no real solution for $$x$$ in this case. (Note: We are looking for real solutions. If we were looking for complex solutions, then raising both sides to the power of 4 would give x = 16, but $$\sqrt[4]{16}$$ is defined as the positive real root, which is 2, not -2. So 16 is not a solution either.) Now, let's check our solution $$x = 625$$ in the original equation: $$\sqrt{625} - 3\sqrt[4]{625} - 10 = 0$$ $$25 - 3(5) - 10 = 0$$ $$25 - 15 - 10 = 0$$ $$0 = 0$$ So, $$x = 625$$ is a valid solution. Answer: 625