# Continuity and roots of a polynomial (Tomato subjective 75)

problem: Show that there is at least one value of $${x}$$ for which $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$$
solution: $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$$
$${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1}$$ is a continuous function. Now if we can chose $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1}$$ takes both negative & positive numbers then the $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 = 0}$$ have a solution.

For $${\displaystyle{x = {\frac{1}{64}}}}$$, $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 < 0}$$ & for $${x = 64}$$, $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 > 0}$$.
So by intermediate value theorem we can say that $${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} – 1 = 0}$$ has a solution for some real $${x}$$.