Given that one root of an equations #3x^2+kx+100=0# is three times the other root. What is #k#?

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Answer 1 · 2017-06-24T16:13:43

#k=+-40#

#for any quadratic

#ax^2+bx+c=0#

if the roots are #alpha, beta#

then

#alpha+beta=-b/a#

#alphabeta=c/a#

in this case we have;

#3x^2+kx+100=0#

and one root is three times the other

so we have

the roots are

#alpha, 3alpha#

#=>alpha+3alpha=>4alpha=-k/3---(1)#

#alphaxx3alpha=100/3#

#=>3alpha^2=100/3--(2)#

from#" "(1)" "alpha=-k/12#

substitute into #(2)#

#3(-k/12)^2=100/3#

#3(k^2/144)=100/3#

#k^2=100/cancel(3)xxcancel(144)^16/cancel(3)#

#k^2=1600#

#k=+-sqrt1600=+-40#