Suppose that a is a nonzero constant for which the equation ax^2 + 20x + 7 = 14x + 9 has only one solution. Find this solution.
+1622 First we must simplify this equation.
We subtract (14x + 9) from both sides of the equation to get the quadratic \(ax^2 + 6x - 2\) = 0
Then we apply the quadratic formula \(-b {+\over} \sqrt{b^2 - 4ac}\over2a\) where a is the coefficient of x^2, b is the coefficient of x, and c is the constant.
Plugging it in, we get -3 + or - \(\sqrt{11}\), the nonzero constant (for which 'a' has one solution) will most likely be the positive solution.
\(a = \sqrt{11} - 3\)
