Question

Given (8x+3k)(9x−2)=0, the product of all solutions is (−3). Find k.

Answer 1

Answer:

$k=36$

Step-by-step explanation:

$Given\ (8x+3k)(9x-2)=0\\\\Product\ of\ two\ expressions\ is\ zero\ then\ one\ of\ them\ must\ be\ zero.\\\\If\ (9x-2)=0\\\\Add\ 2\ both\ the\ sides\\\\9x-2+2=0+2\\\\9x=2\\\\divide\ both\ sides\ by\ 9.\\\\\frac{9x}{9}=\frac{2}{9}\\\\x=\frac{2}{9}\\\\When\ (8x+3k)=0\\\\8x=-3k\\\\x=\frac{-3k}{8}$

$Solutions\ are\ \frac{2}{9}\ and\ \frac{-3k}{8}\\\\Product\ of\ these\ solutions=-3\\\\\frac{2}{9}\times \frac{-3k}{8}=-3\\\\-\frac{6k}{72}=-3\\\\\frac{2k}{72}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide\ both\ sides\ by\ -3\\\\k=\frac{72}{2}=36$

Other Method:

Product of solutions: Product of the roots of the polynomial $ax^2+bx+c=0$ is given by $\frac{c}{a}$

$(8x+3k)(9x-2)=0\\\\(72x^2+27kx-16x-6k)=0\\\\72x^2+(27k-16)x-6k=0\\\\Here\ a=72,\ b=(27k-16),\ c=-6k\\\\\frac{c}{a}=\frac{-6k}{72}\\\\Product\ is\ given=-3\\\\-\frac{6k}{72}=-3\\\\6k=72\times 3\\\\k=\frac{72\times 3}{6}\\\\k=36$