The equations #2x^2 +3x=4# is rewritten in the form #2(x-h)^2+ q=0#. What is the value of q?

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Answer 1 · 2017-01-07T11:04:37

#q=-41/8#

You would get the equivalent:

1) by subtracting 4:

#2x^2+3x-4=0#

2) by factorizing 2:

#2(x^2+3/2x-2)=0#

3) since

#x^2+3/2x-2=x^2+3/2x color(red)(+9/16-9/16)-2#

and the first three terms are the squared binomial

#(x+3/4)^2#,

you get:

#2((x+3/4)^2-9/16-2)=0#

and then

#2(x+3/4)^2+2(-9/16-2)=0#

where

#q=-9/8-4=-41/8#