Killer 'K'

Algebra Level pending

The smallest value of k for which both the roots of the equation x^2-8kx+16(k^2-k+1)=0 are real, distinct and have values at least 4, is

2 solutions

Ahmed Abdelbasit
Jun 5, 2014

the smaller root of the equation can be calculated by :

$\frac{8k- \sqrt{64k^{2} - 64(k^{2} - k +1)} }{2}$ & $\frac{8k+ \sqrt{64k^{2} - 64(k^{2} - k +1)} }{2}$

to have real distinct roots .. the value under $\sqrt{.... }$ must be positive

so : $64k -64$ $>$ $0$

then : $k$ $>$ $1$

so.. try $k = 2$ ... this gives the roots : $4 , 12$ #

Bleach Byakuya
Jun 19, 2014

Man ! k is not necessary an integer right ?

using b^2 -4ac > 0 , 64k^2 - 64(k^2 -k +1 ) > 0 k > 1

Bleach Byakuya - 6 years, 11 months ago