Question

If () and () are the roots of the quadratic equation (2x^2-3x-1=0), then find the value of (^4+^4).

• A. ( 9716 )
• B. ( 16116 )
• C. ( 14516 )
• D. ( 15316 )

Hint:

For calculating symmetric power functions of roots: 1. First write down (S_1 = + ) and (P = ). 2. Calculate (S_2 = ^2 + ^2 = S_1^2 - 2P). 3. Calculate (S_4 = ^4 + ^4 = S_2^2 - 2P^2). Keeping this structured sequence in mind reduces any execution mistakes down to zero!

Answer 1

The Correct Option is B

Step 1: Recall the root relations.
For $Ax^2 + Bx + C = 0$ with roots $\alpha$ and $\beta$, the sum of roots is $\alpha + \beta = -\frac{B}{A}$ and the product is $\alpha\beta = \frac{C}{A}$. These come straight from comparing coefficients.

Step 2: Read off the values.
Here $A = 2$, $B = -3$, $C = -1$. So $\alpha + \beta = \frac{3}{2}$ and $\alpha\beta = -\frac{1}{2}$.

Step 3: Build $\alpha^2 + \beta^2$.
Use $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$. \[ \left(\tfrac{3}{2}\right)^2 - 2\left(-\tfrac{1}{2}\right) = \tfrac{9}{4} + 1 = \tfrac{13}{4} \]

Step 4: Step up to $\alpha^4 + \beta^4$.
Use $\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2$. The idea is the same, square the lower sum then take off twice the product squared.

Step 5: Plug in the numbers.
\[ \left(\tfrac{13}{4}\right)^2 - 2\left(-\tfrac{1}{2}\right)^2 = \tfrac{169}{16} - 2\cdot\tfrac{1}{4} \]

Step 6: Finish the arithmetic.
$2 \cdot \frac{1}{4} = \frac{1}{2} = \frac{8}{16}$, so $\frac{169}{16} - \frac{8}{16} = \frac{161}{16}$. Therefore \[ \boxed{\dfrac{161}{16}} \]