Given the equation #3x^2 +2x +k =0#, how do you state the sum and product of the roots?

1 Answer
Jan 2, 2017

Answer:

The sum of the roots is #-2/3# and their product is #k/3#

Explanation:

Given:

#3x^2+2x+k = 0#

Divide through by #3# to get the monic quadratic equation:

#x^2+2/3x+k/3 = 0#

Note that if this has roots #r_1# and #r_2# then:

#0 = (x-r_1)(x-r_2) = x^2-(r_1+r_2)x+r_1 r_2#

So equating coefficients, we find that the sum of the roots is #-2/3# and the product of the roots is #k/3#

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Footnote

Consider what happens for a polynomial of degree #n#.

When we multiply binomials like this:

#(x-r_1)(x-r_2)...(x-r_n)#

#=x^n#

#-(r_1+r_2+...+r_n)x^(n-1)#

#+(r_1r_2+r_1r_3+...+r_1r_n+r_2r_3+...+r_(n-1)r_n)x^(n-2)#

#-(r_1r_2r_3+...+r_(n-2)r_(n-1)r_n)x^(n-3)#

#+...+-r_1r_2...r_n#

The coefficients of the resulting polynomial are called the elementary symmetric polynomials in #r_1, r_2,..., r_n#

Any symmetric polynomial in #r_1, r_2, ..., r_n# can be expressed in terms of these elementary ones.

For example, when #n = 2# we can express #r_1^2+r_2^2# in terms of the elementary symmetric polynomials #r_1+r_2# and #r_1r_2#:

#r_1^2+r_2^2 = (r_1+r_2)^2 - 2(r_1r_2)#