Question

Suppose that a is a nonzero constant for which the equation ax^2+20x+7=0 has only one solution. find this solution.

Answer 1

$\Delta=0\\\\ \Delta=20^2-4\cdot a\cdot7=400-28a\\ 400-28a=0\\ 28a=400\\ a=\dfrac{400}{28}=\dfrac{100}{7}$