Question

If the sum of the zeroes of the polynomial (p(x)=2x^3-3kx^2+4x-5) is 6, then the value of k is

• A. (5)
• B. (-5)
• C. (4)
• D. (-4)

Hint:

Vieta's formula signs always alternate systematically starting from the second term! * Sum of roots taken 1 at a time (( )) = (-2nd coefficient1st coefficient) * Sum of roots taken 2 at a time (( )) = (+3rd coefficient1st coefficient) * Product of roots taken 3 at a time (()) = (-4th coefficient1st coefficient) Memorizing this sign-alternation prevents common negative sign errors during fast-paced exam environments.

Answer 1

The Correct Option is C

Step 1: Recall the cubic root rule.
For $Ax^3 + Bx^2 + Cx + D = 0$ with roots, the sum of the roots equals $-\frac{B}{A}$. This comes from matching coefficients in Vieta's formulas.

Step 2: Read the coefficients.
The polynomial is $2x^3 - 3kx^2 + 4x - 5$. So $A = 2$ and $B = -3k$.

Step 3: Write the sum of zeroes.
\[ \text{Sum} = -\frac{B}{A} = -\frac{-3k}{2} = \frac{3k}{2} \]

Step 4: Use the given sum.
We are told the sum of the zeroes is $6$. \[ \frac{3k}{2} = 6 \]

Step 5: Clear the fraction.
Multiply both sides by $2$ to get $3k = 12$.

Step 6: Solve for $k$.
Divide by $3$, giving $k = 4$. So the required value is \[ \boxed{4} \]