The coefficients $a_{2} and a_{1}$ $a_2 and a_1$ of a 2nd order polynomial $a_{2} x^{2} + a_{1} x + a_{0} = 0$ $a_2x^2+a_1x+a_0=0$ 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

$a_{2} x^{2} + a_{1} x + a_{0} = 0$ $a_2x^2+a_1x+a_0=0$

$a_{2} = 3$ $a_2=3$

$a_{1} = 5$ $a_1=5$

one root is $\frac{1}{3}$ $1/3$

for a quadratic if $α, β$ $alpha, beta$ are the roots then

$α + β = - \frac{a_{1}}{a_{2}}$ $alpha+beta=-a_1/a_2$

$α β = \frac{a_{0}}{a_{2}}$ $alphabeta=a_0/a_2$

from the information given:

let $α = \frac{1}{3}$ $alpha=1/3$

$\frac{1}{3} + β = - \frac{5}{3}$ $1/3+beta=-5/3$

$β = - \frac{5}{3} - \frac{1}{3} = - \frac{6}{3} = - 2$ $beta=-5/3-1/3=-6/3=-2$