The coefficients #a_2 and a_1# of a 2nd order polynomial #a_2x^2+a_1x+a_0=0# are 3 and 5 respectively. One solution of the polynomial is 1/3. Determine the other solution?

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Answer 1 · 2017-01-19T08:21:17

-2

#a_2x^2+a_1x+a_0=0#

#a_2=3#

#a_1=5#

one root is #1/3#

for a quadratic if #alpha, beta# are the roots then

#alpha+beta=-a_1/a_2#

#alphabeta=a_0/a_2#

from the information given:

let #alpha=1/3#

#1/3+beta=-5/3#

#beta=-5/3-1/3=-6/3=-2#