If 2 + 3i is one of the roots of the equation | vedclass

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(B) Since the coefficients of the polynomial $2x^3 - 9x^2 + kx - 13 = 0$ are real,complex roots must occur in conjugate pairs.Given one root is $\alph

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exists and is equal to $\frac{1}{2}.$

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(B) Since the coefficients of the polynomial $2x^3 - 9x^2 + kx - 13 = 0$ are real,complex roots must occur in conjugate pairs.Given one root is $\alpha = 2 + 3i,$ the other complex root must be $\beta = 2 - 3i.$Let the real root be $\gamma.$According to the relation between roots and coefficients for a cubic equation $ax^3 + bx^2 + cx + d = 0,$ the product of the roots is given by $\alpha \beta \gamma = -\frac{d}{a}.$Here,$a = 2$ and $d = -13,$ so the product of the roots is $-\frac{-13}{2} = \frac{13}{2}.$Substituting the known roots: $(2 + 3i)(2 - 3i) \gamma = \frac{13}{2}.$$(2^2 + 3^2) \gamma = \frac{13}{2}.$$(4 + 9) \gamma = \frac{13}{2}.$$13 \gamma = \frac{13}{2}.$$\gamma = \frac{1}{2}.$Thus,the real root exists and is equal to $\frac{1}{2}.$

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation:

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