A quadratic space over an algebraically closed field is isotropic

Let $F$ be an algebraically closed field, and let $(V,f)$ be a quadratic space over $F$. How can one show that if $\dim V \geq 2$ then it is an isotropic space?

Thanks.

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Choose an arbitrary basis of $V$ denoted by $\{v_i\}$. Consider the equation
$$f(v_1 + k v_2) = 0$$
which is a quadratic polynomial in $k$. (Define $q_1 = f(v_1)$, $q_2 = f(v_2)$, and $q_{12} = f(v_1 + v_2)$. You can show that $f(v_1 + k v_2) = q_1 + k^2 q_2 + k( q_{12} - q_1 - q_2 )$.)
Since $F$ is algebraically closed, you can solve for $k$ a root. Then the vector $v = v_1 + k v_2$ (which is non-zero since $v_1,v_2$ are linearly independent) is an isotropic vector.