Question 181796
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*[tex \LARGE \text{          }\math x^2 +2x + {3 \over 4} = 0]


Note that


*[tex \LARGE \text{          }\math ({1 \over 2})({3 \over 2}) = {3 \over 4}]


Which is the constant term of your equation and 


*[tex \LARGE \text{          }\math ({1 \over 2}) + ({3 \over 2}) = {4 \over 2} = 2]


Which is the coefficient on the 1st degree term in your equation, so


*[tex \LARGE \text{          }\math x^2 +2x + {3 \over 4} = 0\ \ \Rightarrow\ \ (x + {1 \over 2})(x + {3 \over 2}) = 0]


Verification by using FOIL is left as an exercise for the student.


Using the Zero Product Rule:


*[tex \LARGE \text{          }\math (x + {1 \over 2} )(x + {3 \over 2}) = 0\ \ \Rightarrow\ \ x + {1 \over 2} = 0 \text{ or } x + {3 \over 2} = 0\ \ \Rightarrow\ \ x = -{1 \over 2} \text{ or } x = -{3 \over 2}]


And the graph verifies it:


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
circle(-1/2,0,.08),
circle(-3/2,0,.08),
graph(
500, 500, -5, 5, -5, 5,
y = x^2 +2x + 3/4
))}}}


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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