Why does the same method of completing the square fail when the value of "a" does not equal 1 for a quadratic? ax^2 + bx + c

Answer 1

the method of " completing the square " works for all values of a except 0.
In that case the ax^2 term would drop out, and you would not have a quadratic.
I don't know what "same method" you are referring to.
Here is the standard method applied to a specific case

e.g let a = 3/4, such as

y = (3/4)x^2 + (5/8)x + 2/7
= (3/4)(x^2 + (5/6)x + .... - ...) + 2/7
= (3/4)(x^2 + (5/6)x + 25/144 - 25/144) + 2/7
= (3/4)( (x + 5/12)^2 - 25/144) + 2/7
= (3/4)(x+5/12)^2 - 25/192 + 2/7
= (3/4)(x + 5/12)^2 + 209/1344

Answer 2

Oh, the "same method" is probably referring to this:

4. Write each quadratic in vertex form by completing the square. Then identify the vertex and solve the quadratic to find the x intercepts.