Question 232351
The general form of the quadratic equation is:
{{{ax^2+bx+c=0}}}
Your equation is
{{{p^2x^2 + (p^2-q^2)x-q^2=0}}}
or 
{{{(p^2)x^2 + (p^2-q^2)x + (-q^2)=0}}}<br>
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^2}}} for "a", {{{(p^2-q^2)}}} for "b" and {{{-q^2}}} for "c". And we can substitute these into the quadratic formula,
{{{x = (-b +- sqrt(b^2 -4ac))/(2a)}}} giving us:
{{{x = (-(p^2-q^2) +- sqrt((p^2-q^2)^2 - 4(p^2)(-q^2)))/(2(p^2))}}}
Now we just simplify as much as possible:
{{{x = (-p^2+q^2 +- sqrt(p^4- 2p^2q^2 + q^4 + 4p^2q^2))/(2p^2)}}}
{{{x = (-p^2+q^2 +- sqrt(p^4+ 2p^2q^2 + q^4))/(2p^2)}}}
{{{x = (-p^2+q^2 +- sqrt((p^2+ q^2)^2))/(2p^2)}}}
{{{x = (-p^2+q^2 +- (p^2+ q^2))/(2p^2)}}}
(Note: Normally {{{sqrt(x^2) = abs(x)}}}. But since {{{p^2 + q^2}}} 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 = (-p^2+q^2 + (p^2+ q^2))/(2p^2)}}}
or
{{{x = (-p^2+q^2 - (p^2+ q^2))/(2p^2)}}}
And we will finish simplifying them separately.
{{{x = (2q^2)/(2p^2) = q^2/p^2}}}
or
{{{x = (-2p^2)/(2p^2) = -1}}}