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

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(B) Given the equation $2x^3 - 9x^2 + kx - 13 = 0$ with real coefficients,complex roots must occur in conjugate pairs.Since $2 + 3i$ is a root,its con

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

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(B) Given the equation $2x^3 - 9x^2 + kx - 13 = 0$ with real coefficients,complex roots must occur in conjugate pairs.Since $2 + 3i$ is a root,its conjugate $2 - 3i$ must also be a root.Let the roots be $\alpha = 2 + 3i,$ $\beta = 2 - 3i,$ and $\gamma$ be the real root.According to the relationship 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 $\alpha \beta \gamma = -\frac{-13}{2} = \frac{13}{2}.$Calculating the product of the complex roots: $\alpha \beta = (2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 + 9 = 13.$Substituting these into the product formula: $13 \cdot \gamma = \frac{13}{2}.$Solving for $\gamma,$ we get $\gamma = \frac{1}{2}.$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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