How do you find the value of #k,# given that one solution of the equation #x^2-kx+15=0# is #1/2#?

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Answer 1 · 2017-07-15T21:46:20

See a solution process below:

To find #k#, substitute #color(red)(1/2)# for each occurrence of #color(red)(x)# in the equation and solve for #k#:

#color(red)(x)^2 - kcolor(red)(x) + 15 = 0# becomes:

#color(red)(1/2)^2 - (k * color(red)(1/2)) + 15 = 0#

#1/4 - 1/2k + 15 = 0#

#-1/2k + 15 + 1/4 = 0#

#-1/2k + (4/4 xx 15) + 1/4 = 0#

#-1/2k + 60/4 + 1/4 = 0#

#-1/2k + 61/4 = 0#

#-1/2k + 61/4 - color(red)(61/4) = 0 - color(red)(61/4)#

#-1/2k + 0 = -61/4#

#-1/2k = -61/4#

#color(red)(-2) xx -1/2k = color(red)(-2) xx 61/(-4)#

#1k = cancel(color(red)(-2)) xx 61/(color(red)(cancel(color(black)(-4)))2)#

#k = 61/2#