In this problem, assume that $x$ is positive (so we can take its square root).

Enter (without the quotation marks) "$==>$" if the left equation implies the right, "$<==$" if the right equation implies the left, "$<==>$" if either equation implies the other, and "$><$" if neither equation implies the other.

$x-5\sqrt{x} +6 = 0$ $x^2-5x+6 = 0$.

$x-5\sqrt{x} +6 = 0$ $5\sqrt{x} = x+ 6$.

$5\sqrt{x} = x+ 6$ $25x = x^2 + 12 x + 36$.

This last equation is an ordinary quadratic equation. Solve it to answer the remaining question: The smaller solution of $x-5\sqrt{x} +6 = 0$ is $x=$, and the larger is $x=$.