1)Solve the following quadratic equations
a)x2 + 4x + 3 =
b) x2 - 5x + 6
c) x2 + 2x – 3
d)x2 – 2x – 8
2)Give the values of m and c
a)y = 2x + 3
b) y = 4x – 2
c)y = 2x – 5
d)y = 1/2x – 8
e) 3y = 6x – 9
f) 1/2y = 2x + 3
These are the answers I came up for question no #2. Comparing the given equations with the standard format of
y = mx + c
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a)
y = 2x + 3
m = 2 and c = 3
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b)
y = 4x - 2
m = 4 and c = -2
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c)
y = 2x - 5
m = 2 and c = -5
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d)
y = 1/2x - 8
m = 1/2 and c = -8
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e)
3y = 6x - 9
rearranging y = 2x -3
m = 2 and c = -3
---------------------------------------------------------------
f)
1/2y = 2x + 3
rearranging y = 4x + 6
m = 4 and c = 6
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These are expressions; having no "equals" signs in them, they cannot be "equations". Please reply with clarification.
For the first four, the equations are already written in the form "y = mx + c". Copy off the values of "m" and "c".
For the last two, solve for "y=", and then copy off the values.
Oh that makes sense. Well if these equations are set to zero than an easier method (if you are comfortable factoring) is to do so:
factors to . And thus the resultant equation:
we can solve by taking any of the two terms and setting it individually equal to zero:
Using the quadratic equation also works, but might be a bit of a waste on equations that factor nicely.
On the second set, "Give the values of M and C" I'm assuming refers to:
;
where is the slope of our equation and is our y-intercept and the equation is the slope-intercept form of a line.
Thusly, arranging each of the equations a-f and putting them in slope-intercept form will yield our M and C.