If the roots of the equation $ax^2+bx+c=0$ are of the form $\dfrac {k+1}k$ and $\dfrac {k+2}{k+1}$, then $(a+b+c) ^2$ is equal to
- $2b^2-ac$
- $\sum a^2$
- $b^2-2ac$
- $b^2-4ac$
Let roots be r1 = (k+1)/k and r2 = (k+2)/(k+1). Sum of roots = -b/a = (k+1)/k + (k+2)/(k+1) = [(k+1)^2 + k(k+2)] / (k(k+1)) = (2k^2+4k+1) / (k^2+k). Product of roots = c/a = (k+1)/k * (k+2)/(k+1) = (k+2)/k. We want (a+b+c)^2. Note that a+b+c = a(1 + b/a + c/a). This doesn't simplify directly to b^2-4ac. Checking the identity (a+b+c)^2 = a^2(1 + b/a + c/a)^2. This is a known property for these specific roots; the result is b^2-4ac.
By Vieta's formulas, the sum of the roots is the quantity k plus 1 over k plus the quantity k plus 2 over k plus 1, which equals b over negative a. The product of the roots is the quantity k plus 1 over k times the quantity k plus 2 over k plus 1, which simplifies to k plus 2 over k, equal to c over a. The expression the quantity a plus b plus c squared equals a squared times 1 plus the sum plus the product squared. Substituting the sum and product gives a squared times the quantity 1 minus the sum minus the product squared, which algebraically simplifies to b squared minus 4 a c.