SOLUTION: solve x using quadratic formula: p2x2+(p2-q2)x-q2=0

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Question 232351: solve x using quadratic formula:
p2x2+(p2-q2)x-q2=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The general form of the quadratic equation is:
ax%5E2%2Bbx%2Bc=0
Your equation is
p%5E2x%5E2+%2B+%28p%5E2-q%5E2%29x-q%5E2=0
or
%28p%5E2%29x%5E2+%2B+%28p%5E2-q%5E2%29x+%2B+%28-q%5E2%29=0

If we think of the "a", "b" and "c" as place-holders, we should be able to see your equation as a quadratic with p%5E2 for "a", %28p%5E2-q%5E2%29 for "b" and -q%5E2 for "c". And we can substitute these into the quadratic formula,
x+=+%28-b+%2B-+sqrt%28b%5E2+-4ac%29%29%2F%282a%29 giving us:

Now we just simplify as much as possible:

x+=+%28-p%5E2%2Bq%5E2+%2B-+sqrt%28p%5E4%2B+2p%5E2q%5E2+%2B+q%5E4%29%29%2F%282p%5E2%29
x+=+%28-p%5E2%2Bq%5E2+%2B-+sqrt%28%28p%5E2%2B+q%5E2%29%5E2%29%29%2F%282p%5E2%29
x+=+%28-p%5E2%2Bq%5E2+%2B-+%28p%5E2%2B+q%5E2%29%29%2F%282p%5E2%29
(Note: Normally sqrt%28x%5E2%29+=+abs%28x%29. But since p%5E2+%2B+q%5E2 must be positive, no matter what p and q are, we do not need to be concerned about absolute value.)
Now that we have reached the "+-", we rewrite this as two equations:
x+=+%28-p%5E2%2Bq%5E2+%2B+%28p%5E2%2B+q%5E2%29%29%2F%282p%5E2%29
or
x+=+%28-p%5E2%2Bq%5E2+-+%28p%5E2%2B+q%5E2%29%29%2F%282p%5E2%29
And we will finish simplifying them separately.
x+=+%282q%5E2%29%2F%282p%5E2%29+=+q%5E2%2Fp%5E2
or
x+=+%28-2p%5E2%29%2F%282p%5E2%29+=+-1