Question

If $\alpha,\beta,\gamma$ are the roots of \[ x^3-10x^2+7x+8=0, \] match List-I with List-II and choose the correct option.

• A. A-V, B-III, C-I, D-II
• B. A-V, B-III, C-II, D-IV
• C. A-V, B-III, C-II, D-I
• D. A-V, B-II, C-III, D-I

Hint:

For cubic root problems, write down all three Vieta relations before starting calculations.

Answer 1

The Correct Option is C

Step 1: Find the basic symmetric values.
For $x^3-10x^2+7x+8=0$ with roots $\alpha,\beta,\gamma$: sum $\alpha+\beta+\gamma=10$, pair sum $\alpha\beta+\beta\gamma+\gamma\alpha=7$, product $\alpha\beta\gamma=-8.$
Step 2: Entry A, the sum.
$A=\alpha+\beta+\gamma=10.$ This matches list value $V$.
Step 3: Entry B, sum of squares.
$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=100-14=86.$ This matches $III.$
Step 4: Entry C, sum of reciprocals.
$\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{7}{-8}=-\dfrac78.$ This matches $II.$
Step 5: Entry D.
$\dfrac{\alpha}{\beta\gamma}+\dfrac{\beta}{\gamma\alpha}+\dfrac{\gamma}{\alpha\beta}=\dfrac{\alpha^2+\beta^2+\gamma^2}{\alpha\beta\gamma}=\dfrac{86}{-8}=-\dfrac{43}{4}.$ This matches $I.$
Step 6: Read off the matching.
So $A\to V,\ B\to III,\ C\to II,\ D\to I.$ \[ \boxed{A\text{-}V,\ B\text{-}III,\ C\text{-}II,\ D\text{-}I} \]