Question

If one of the zeroes of polynomial is 3x2-8x+2k+1 is seven times the other,find the value of k

Answer 1

Step-by-step explanation:

hope it will help you......

Answer 2

Answer:

$k=\frac{2}{3}$

Step-by-step explanation:

$3x^2-8x+2k+1=0\\comparing \ given \ equation \ with \ general \ quadratic \ equation,\\a=3, \ b=-8, \ c=2k+1\\Solving \ given \ equation \ using \ method \ of \ discriminant,\\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-8)\pm\sqrt{(-8)^2-4(3)(2k+1)}}{2(3)}=\frac{8\pm\sqrt{64-12(2k+1)}}{6}=\frac{8\pm\sqrt{64-24k-12}}{6}\\x=\frac{8\pm\sqrt{52-24k}}{6}=\frac{8\pm\sqrt{4(13-6k)}}{6}=\frac{8\pm2\sqrt{(13-6k)}}{6}\\we \ get \ x_1=\frac{8+2\sqrt{(13-6k)}}{6} \ and \ x_2=\frac{8-2\sqrt{(13-6k)}}{6}$

$given \ that \ x_1=7x_2\\\therefore \ \frac{8+2\sqrt{13-6k}}{6}=7[\frac{8-2\sqrt{13-6k}}{6}]\\\therefore \ 8+2\sqrt{13-6k}=7[8-2\sqrt{13-6k}]\\\therefore \ 8+2\sqrt{13-6k}=56-14\sqrt{13-6k}\\\therefore \ 2\sqrt{13-6k}+14\sqrt{13-6k}=56-8\\\therefore \ 16\sqrt{13-6k}=48\\\therefore \ \sqrt{13-6k}=\frac{48}{13}\\\therefore \ \sqrt{13-6k}=3\\\therefore \ (\sqrt{13-6k})^2=3^2\\\therefore \ 13-6k=9\\\therefore \ 6k=13-9\\\therefore \ 6k=4\\\\\therefore \ k=\frac{4}{6}\implies k=\frac{2}{3}$