Can someone work out this Parabola question?

Determine the exact values of 'a' such that the function y = x^2 + 3ax +5 intersects the x axis at only one point. 10 points for the best answer^^





please explain and show working out





ty in advanceCan someone work out this Parabola question?
y = x^2 + 3ax + 5





Since the coefficient of the x^2 term %26gt; 0, then this parabola opens upward. Because it has two sides that extend upward, the only way this parabola can have only one x intercept is to make the vertex the x intercept.





To calculate the x intercept, we make y = 0 for this equation.


y = x^2 + 3ax + 5


0 = x^2 + 3ax + 5


The next step is to factor the right hand side. However, since we want to limit x to just one value, the factors must be equal. The only way to obtain equal factors is to complete the square on the right hand side. Since the last term is 5, the factors will include the square root of 5.


0 = (x + 鈭?)(x + 鈭?)


this limits x to just one value (-鈭?)


next we multiply both factors


0 = x^2 + (2鈭?)x + 5


Unfortunately the coefficient of the middle term (2鈭?) does not match the coefficient of the middle term in the original equation above (3a). So we set up an equality equation using both of these middle terms, then solve for a.


2鈭? = 3a


a = 2鈭?/3 (exact value)


a = 1.49 (rounded off value)