Asked by Henry
Show that the root equation (x-a)(x-b)=k^2 are always real numbers
Answers
Answered by
Damon
x^2 - (a+b) x + a b - k^2 = 0
x = [(a+b) +/- sqrt { (a+b)^2 -4(ab-k^2)} ]/2
if
(a+b)^2 -4(ab-k^2)
is positive, there is no complexity
= a^2 + 2 ab +b^2 -4ab + 4k^2
= a^2 -2ab + b^2 + 4 k^2
= (a-b)^2 + 4 k^2
BOTH of those terms are positive being squares of real numbers, whew we did it !
x = [(a+b) +/- sqrt { (a+b)^2 -4(ab-k^2)} ]/2
if
(a+b)^2 -4(ab-k^2)
is positive, there is no complexity
= a^2 + 2 ab +b^2 -4ab + 4k^2
= a^2 -2ab + b^2 + 4 k^2
= (a-b)^2 + 4 k^2
BOTH of those terms are positive being squares of real numbers, whew we did it !
Answered by
manu
show that the root of the equation x^2+2x=2x(2a+2b+1)(2a+2b-1) are integers if a and b are integer
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