Find the non-zero value of $a$ such that the quadratic equation $ax^2+8x+4=6x-40$ has only one solution.
+177 The given quadratic equation is \(ax^2 + 8x + 4 = 6x - 40\). First, simplify it to standard form.
\(ax^2 + 8x + 4 = 6x - 40 \\ ax^2 + 2x + 44 = 0\)
Now, we can use the discriminant of a quadratic, which gives information about the number of solutions. In order for the quadratic to have only one solution, the discriminant should be equal to 0.
\(D = 0 \\ b^2 - 4ac = 0 \\ 2^2 - 4 * a * 44 \\ 4 - 176a = 0 \\ 176a = 4 \\ a = \frac{4}{176} \\ a = \frac{1}{44}\)
+177 The given quadratic equation is \(ax^2 + 8x + 4 = 6x - 40\). First, simplify it to standard form.
\(ax^2 + 8x + 4 = 6x - 40 \\ ax^2 + 2x + 44 = 0\)
Now, we can use the discriminant of a quadratic, which gives information about the number of solutions. In order for the quadratic to have only one solution, the discriminant should be equal to 0.
\(D = 0 \\ b^2 - 4ac = 0 \\ 2^2 - 4 * a * 44 \\ 4 - 176a = 0 \\ 176a = 4 \\ a = \frac{4}{176} \\ a = \frac{1}{44}\)
