Multiple choice

If roots of the equation $ { ax }^{ 2 }+bx+c=0$ are $ \dfrac { \alpha }{ \alpha -1 } $ and $ \dfrac { \alpha+1 }{ \alpha } $, then $ { \left( a+b+c \right) }^{ 2 }={ b }^{ 2 }+4ac$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots are r1 = a/(a-1) and r2 = (a+1)/a. Sum = r1+r2 = -b/a, Product = r1*r2 = c/a. The identity (a+b+c)^2 = b^2+4ac is not generally true for these roots. Testing with a simple case: if a=1, b=-2, c=0, roots are 2 and 2, but the equation doesn't hold.

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