If $\alpha$ and $\beta$ are roots of $ax^2+bx+c=0$, then equation whose roots are $\displaystyle \alpha +\frac{1}{\beta }, \beta +\frac{1}{\alpha }$, is
- $\displaystyle acx^{2}-b\left ( a-c \right )x+\left ( a+c \right )^{2}= 0.$
- $\displaystyle acx^{2}+b\left ( a+b \right )x-\left ( bc \right )^{2}= 0.$
- $\displaystyle acx^{2}+b\left ( a+c \right )x+\left ( a+c \right )^{2}= 0.$
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None of these
Given alpha+beta = -b/a and alpha*beta = c/a. The new roots are S = alpha+1/beta + beta+1/alpha = (alpha+beta) + (alpha+beta)/(alpha*beta) = -b/a + (-b/a)/(c/a) = -b/a - b/c = -b(a+c)/(ac). The product P = (alpha+1/beta)(beta+1/alpha) = alpha*beta + 1 + 1 + 1/(alpha*beta) = c/a + 2 + a/c = (c^2+2ac+a^2)/(ac) = (a+c)^2/(ac). The equation is x^2 - Sx + P = 0, which leads to acx^2 + b(a+c)x + (a+c)^2 = 0.
Let the new roots be P = α + 1/β and Q = β + 1/α. Since α and β are roots of ax^2 + bx + c = 0, we know the sum α + β = -b/a and the product αβ = c/a. The sum of the new roots is P + Q = α + β + 1/α + 1/β = (α + β) + (α + β)/(αβ) = (-b/a) + (-b/a)/(c/a) = -b(a + c)/(ac). The product of the new roots is PQ = (α + 1/β)(β + 1/α) = αβ + 1 + 1 + 1/(αβ) = c/a + 2 + a/c = (a + c)^2/(ac). Substituting these into x^2 - (P+Q)x + PQ = 0 and multiplying throughout by ac gives acx^2 + b(a + c)x + (a + c)^2 = 0.