Expert: Ahmed Salami - 1/28/2006

Question
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Followup To
Question -
Instructions from textbook: Advanced Mathematical Concepts, Pre-calculus with Applications, by Glencoe McGrawl Hill. Copyright 1994


Use the quadratic equation to solve each equation for "Y" Then use a graphing calculator to draw the graph.

2x^2 + 4xy + 2y^2 + 2times(square root 2)times x + (-)2times(square root 2)times y = -12.

They want the student to use the eqation ay^2+by+c to be used to find a b and c and plug them in the quadratic formula. I have attempted the problem and achieved   2y^2 + (4x-2times(squareroot2))y + (2x^2 + 2times(squareroot2)timesx + 12)
What is does the graph look like? What is "Y" Most importantly, I need the graph, and the solution. If you have access to a solution manuel for this texbook, please use it. If you do not, feel free to ask me anymore questions at my email sunnysunsun@sbcglobal.net

Answer
Hi Georgia,
Sorry for the time it took. Technical problem with the site.
If 2x^2 + 4xy + 2y^2 + 2(sqrt2)x - 2(sqrt2)y = -12
dividing through by 2
x^2 + 2xy + y^2 + (sqrt2)x - (sqrt2)y + 6 = 0
rearranging
y^2 + 2xy - (sqrt2)y + x^2 + (sqrt2)x + 6 = 0
y^2 + [2x - (sqrt2)]y + [x^2 + (sqrt2)x + 6] = 0
comparing with ay^2 + by + c = 0
a = 1
b = 2x - (sqrt2)
c = x^2 + (sqrt2)x + 6
We now use the quadratic formula
y = [-b +or- sqrt(b^2 - 4ac)]/2a
simplifying b^2 - 4ac, we have
[2x - (sqrt2)]^2 - 4(1)[x^2 + (sqrt2)x + 6] =
[4x^2 - 4(sqrt2)x + 2] - [4x^2 + 4(sqrt2)x + 24] =
-8(sqrt2)x - 22
Therefore,
y = (sqrt2 - 2x) +or- sqrt[-8(sqrt2)x - 22] / 2
The graph would most likely be a parabola.
As a privilege, i'd let you have my email address.
sa_ahmed2001@yahoo.com
I hope i have helped. You can always get back to me.
Regards.