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) 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

If $e^{it} = \cos t + i \sin t$ and $e^{-it} = \cos t - i \sin t$,then $\cosh(x + iy) - \cosh(x - iy) =$

Let the set of all values of $k \in R$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z \in C$,has at least one solution,be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

The $A + iB$ form of $\frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i\tan v)}$ is

If $z$ is a complex number satisfying $|z^3+z^{-3}| \leq 2$,then the maximum possible value of $|z+z^{-1}|$ is

If $\sqrt{-3-4 i}=re^{i \theta}$,then $r^2 \tan \theta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module