Answer:
[tex]k=36[/tex]
Step-by-step explanation:
[tex]Given\ (8x+3k)(9x-2)=0\\\\Product\ of\ two\ expressions\ is\ zero\ then\ one\ of\ them\ must\ be\ zero.\\\\If\ (9x-2)=0\\\\Add\ 2\ both\ the\ sides\\\\9x-2+2=0+2\\\\9x=2\\\\divide\ both\ sides\ by\ 9.\\\\\frac{9x}{9}=\frac{2}{9}\\\\x=\frac{2}{9}\\\\When\ (8x+3k)=0\\\\8x=-3k\\\\x=\frac{-3k}{8}\\\\[/tex]
[tex]Solutions\ are\ \frac{2}{9}\ and\ \frac{-3k}{8}\\\\Product\ of\ these\ solutions=-3\\\\\frac{2}{9}\times \frac{-3k}{8}=-3\\\\-\frac{6k}{72}=-3\\\\\frac{2k}{72}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide\ both\ sides\ by\ -3\\\\k=\frac{72}{2}=36[/tex]
Other Method:
Product of solutions: Product of the roots of the polynomial [tex]ax^2+bx+c=0[/tex] is given by [tex]\frac{c}{a}[/tex]
[tex](8x+3k)(9x-2)=0\\\\(72x^2+27kx-16x-6k)=0\\\\72x^2+(27k-16)x-6k=0\\\\Here\ a=72,\ b=(27k-16),\ c=-6k\\\\\frac{c}{a}=\frac{-6k}{72}\\\\Product\ is\ given=-3\\\\-\frac{6k}{72}=-3\\\\6k=72\times 3\\\\k=\frac{72\times 3}{6}\\\\k=36[/tex]
