Asked by Kaur
the roots of the equation
(ab-1)x²+(a+b)x-ab=0 are always
a) real
b) equal
c) complex
d) cannot be determined
Answers
Answered by
Reiny
You have to look at the discriminant b^2 - 4ac of the general equation formula
(a+b)^2 - 4(ab - 1)(-ab)
= a^2 + 2ab + b^2 + 4a^2 b^2 - 4ab
= a^2 - 2ab + b^2 + 4a^2b^2
= (a-b)^2 + 4a^2b^2
both terms clearly are both positive, so the discriminant is positive
Therefore you will have real roots
(a+b)^2 - 4(ab - 1)(-ab)
= a^2 + 2ab + b^2 + 4a^2 b^2 - 4ab
= a^2 - 2ab + b^2 + 4a^2b^2
= (a-b)^2 + 4a^2b^2
both terms clearly are both positive, so the discriminant is positive
Therefore you will have real roots
Answered by
Kaur
Thank you