Question
given that a,b,c are real numbers and c is bigger than 0, prove that the roots of the equation (x-a)(x-b)=c are real.
Answers
Steve
expand and collect to get
x^2 - (a+b)x + ab-c = 0
The discriminant is
(a+b)^2 - 4(ab-c)
= a^2+2ab+b^2 - 4ab + c
= (a-b)^2 + c
Since c>0, the discriminant is always positive, and the roots are real.
x^2 - (a+b)x + ab-c = 0
The discriminant is
(a+b)^2 - 4(ab-c)
= a^2+2ab+b^2 - 4ab + c
= (a-b)^2 + c
Since c>0, the discriminant is always positive, and the roots are real.