Question 160427
From {{{x^2 + px -q^2 = 0}}} we can see that {{{a=1}}}, {{{b=p}}}, and {{{c=-q^2}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=p^2-4(1)(-q^2)}}} Plug in {{{a=1}}}, {{{b=p}}}, and {{{c=-q^2}}}



{{{D=p^2+4q^2}}} Multiply 



So for <b>any</b> value of "p", the value of {{{p^2}}} is nonnegative. So {{{p^2}}} is <b>always</b> nonnegative (ie it is zero or a positive number).


Also for <b>any</b> value of "q", the value of {{{q^2}}} is nonnegative. So by extension, {{{4q^2}}} is <b>always</b> nonnegative.



So this means that {{{p^2+4q^2}}} is <b>always</b> nonnegative which means that {{{p^2+4q^2>=0}}} and {{{D>=0}}}. Since the discriminant D is greater than or equal to zero, this means that the equation {{{x^2 + px -q^2 = 0}}} will have real solutions for any values of "p" and "q"