Question

If 2x+y=0 is one of the lines
represented by 3x^2+kxy+2y^2=0 then
value of k is​

Answer 1

Answer:

k=\frac{11}{2}

Step-by-step explanation:

\documentclass

\begin{documnet}

3\left(-\frac{y}{2}\right)^2+k\left(-\frac{y}{2}\right)y+2y^2=0

\frac{3y^2}{4}-\frac{ky^2}{2}+2y^2=0

\frac{3y^2}{4}-\frac{ky^2}{2}+2y^2-\left(\frac{3y^2}{4}+2y^2\right)=0-\left(\frac{3y^2}{4}+2y^2\right)

\frac{3y^2}{4}-\frac{ky^2}{2}+2y^2-\left(\frac{3y^2}{4}+2y^2\right)=0-\left(\frac{3y^2}{4}+2y^2\right)

-\frac{ky^2}{2}=-2y^2-\frac{3y^2}{4}

2\left(-\frac{ky^2}{2}\right)=-2\cdot \:2y^2-2\cdot \frac{3y^2}{4}

-ky^2=-4y^2-\frac{3y^2}{2}

\frac{-ky^2}{-y^2}=-\frac{4y^2}{-y^2}-\frac{\frac{3y^2}{2}}{-y^2};\quad \:y\ne \:0

k=\frac{11}{2};\quad \:y\ne \:0

\end{document}