Given that one root is 3 times the others for the quadratic equation #3x^2-2x+p=0#, find (a) the value of #p# and (b) the two roots ?

1 Answer
iceman
May 4, 2018

#p=1/4, x=1/6, 1/2#

Explanation:

Let the roots be #r, 3r#:

#a(x-r)(x-3r)=0#

#ax^2-4arx+3ar^2=0#

#3x^2-2x+p=ax^2-4arx+3ar^2#

#a=3#

#3x^2-2x+p=3x^2-12rx+9r^2#

#12r=2#

#r=1/6=># #3r=1/2#

#p=9*1/(36)=1/4#

Check:

#3x^2-2x+1/4=0#
#3(x^2-(2/3)x)=-1/4#
#3(x^2(-2/3)x+(1/3)^2)=1/3-1/4=1/12#
#(x-1/3)^2=1/36#
#x-1/3=+-1/6#
#x=1/3+-1/6#
#x=1/3-1/6=2/6-1/6=1/6#
#x=1/3+1/6=2/6+1/6=3/6=1/2#

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