Question

x^2-7x+12=0 solve, completing the square

Answer 1

3,4

Answer 2

can u show how?

Answer 3

(x-4)(x+3)=0

Answer 4

\[ x^2-7x+12=0 \Rightarrow (x-4)(x+3)=0 \Rightarrow x =3,4\]

Answer 5

Right answer but not by completing the square:

\[(x-7/2) ^{2} - 1/4 = 0\]
(x-7/2) ^{2} = 1/4
(x-7/2) = sqrt (1/4) = \[\pm1/2\]
x = 7/2 \[\pm\]1/2
x = 6/2 = 3
or x = 8/2 = 4

Answer 6

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