SOLUTION: How to show that the equation x^2 + px -q^2 = 0 are real for any real numbers p & q? Thanks

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Question 160427: How to show that the equation x^2 + px -q^2 = 0 are real for any real numbers p & q? Thanks
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
From x%5E2+%2B+px+-q%5E2+=+0 we can see that a=1, b=p, and c=-q%5E2


D=b%5E2-4ac Start with the discriminant formula.


D=p%5E2-4%281%29%28-q%5E2%29 Plug in a=1, b=p, and c=-q%5E2


D=p%5E2%2B4q%5E2 Multiply


So for any value of "p", the value of p%5E2 is nonnegative. So p%5E2 is always nonnegative (ie it is zero or a positive number).

Also for any value of "q", the value of q%5E2 is nonnegative. So by extension, 4q%5E2 is always nonnegative.


So this means that p%5E2%2B4q%5E2 is always nonnegative which means that p%5E2%2B4q%5E2%3E=0 and D%3E=0. Since the discriminant D is greater than or equal to zero, this means that the equation x%5E2+%2B+px+-q%5E2+=+0 will have real solutions for any values of "p" and "q"