Question

Given the equatiom of (2k-1)x^2 + (k-5)x-6=0 Determine the valuw of k so that one root is 2

Answer 1

If 2 is a root, that means if you plug in 2 for x, you get zero. Plugging in x = 2 yields:\[(2k-1)(2)^2+(k-5)(2)-6=0\Longrightarrow (2k-1)\cdot 4+(k-5)\cdot 2-6=0\]\[\Longrightarrow 8k-4+2k-10-6=0\Longrightarrow \cdots \]