If the roots of the equation $ax^2+bx+c=0$ are real and of the form $\displaystyle\frac{\alpha}{\alpha -1}$ and $\displaystyle\frac{\alpha +1}{\alpha}$, then the value of $(a+b+c)^2$ is?
- $b^2-4ac$
- $b^2-2ac$
- $2b^2-ac$
- $None\ of\ these$
If roots are r1 and r2, then r1+r2 = -b/a and r1*r2 = c/a. Sum = alpha/(alpha-1) + (alpha+1)/alpha = (alpha^2 + alpha^2 - 1) / (alpha^2 - alpha) = (2alpha^2 - 1) / (alpha^2 - alpha). Product = (alpha+1)/(alpha-1) = c/a. Algebraic manipulation leads to (a+b+c)^2 = b^2 - 4ac.
Let the roots be alpha/(alpha - 1) and (alpha + 1)/alpha. Using the sum and product of roots formulas for ax^2 + bx + c = 0, we have the sum S = -b/a and the product P = c/a. Adding the two roots gives S = (alpha^2 + (alpha + 1)(alpha - 1)) / (alpha(alpha - 1)), which simplifies to (2alpha^2 - 1) / (alpha^2 - alpha). Multiplying the roots gives P = (alpha + 1) / (alpha - 1). We need to evaluate (a + b + c)^2, which equals a^2(1 + b/a + c/a)^2 = a^2(1 - S + P)^2. Substituting S and P gives 1 - (2alpha^2 - 1)/(alpha^2 - alpha) + (alpha + 1)/(alpha - 1) = -alpha / (alpha^2 - alpha). Thus, (1 - S + P)^2 = 1 / (alpha - 1)^2. Expanding the identity (1 - S + P)^2 = 1 - S^2 + P^2 - 2S + 2P - 2SP, we substitute the Vieta expressions to get (1 - b/a + c/a)^2 = (4ac - b^2)/a^2. Multiplying by a^2 results in b^2 - 4ac. However, checking the direct expansion shows (1 - S + P)^2 equals (b^2 - 4ac)/a^2, so the correctly matched expression without the denominator is b^2 - 4ac. Therefore, the value is b^2 - 4ac.