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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

Find the number of natural number solutions for the equation $x_1 + x_2 = 100$,such that $x_1$ and $x_2$ are not multiples of $5$.

Number of equations of the form $ax^2 + bx + 1 = 0$ having real roots,where $a, b \in \{1, 2, 3, 4\}$ is-

If $\alpha$ and $\beta$ are two real numbers satisfying $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$,then the value of $\alpha \beta$ is:

If $a + b + c = 0$,$a \neq 0$,$a, b, c \in Q$,then both the roots of the equation $ax^2 + bx + c = 0$ are

If one root of the quadratic equation $2x^{2} + Px + 4 = 0$ is $2$,find the second root and the value of $P$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module