8). The polynomial for which you wish to find zeros is
a^2 - 6 a + 9 = 0
Then I factored
(a-3)(a-3) = 0
a = 3 or a = 3, in other words the vertex is on the x axis
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You could have completed the square (but it is already completed):
a^2 - 6 a = -9
a^2 - 6 a + (6/2)^2 = - 9 +(6/2)^2
(a-3)^2 = 0
a = 3 or 3
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You could have used the quadratic equation
x = [-b +/- sqrt(b^2-4ac) ]/2a
x = [ 6 +/- sqrt (36 -36) ]/2
x = [ 6 +/- 0 ]/2
x = 3
Please help me make sure that I am using my vocabulary words correctly to explain the problem. Thank you!!!
8). The Quadratic Formula a^2 - 6 a + 9 = 0
Then I completed the square-------(a-3)(a-3) = 0
a = 3 or a = 3, in other words the vertex is on the x axis
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