If one of the roots of the equation #x^2+kx=6# is #2#, what is the value of #k#?

Question

8228 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-07T19:37:27

#k = 1#

#x^2 - Sx + P = 0 <=> x^2 + kx - 6 = 0#

#a + b = -k#

#ab = - 6#

#b = 2 => a = -3#

#-3 + 2 = -k#

#-1 = -k#

Answer 2 · 2017-06-07T19:46:26

#k=1#

Let's solve the quadratic for #k#:

#kx=6-x^2#

#k=(6-x^2)/x#

If one root equals #2#, let's plug it on to the equation and find #k#.

#k=(6-(2^2))/2=(6-4)/2=2/2#

#k=1#