The equations \(75x^4 + ax^3 + bx^2 + cx + 12 = 0\)

and \(12x^5 + dx^4 + ex^3 + fx^2 + gx + 75 = 0\)

have a common rational root \(k\) which is not an integer, and which is negative. What is \(k \)

FlyEaglesFly Apr 26, 2019

#1**+3 **

Let k = -p/q, (where p and q are integers), then (x + p/q) will be a factor of both polynomials, so

\(\displaystyle (x+p/q)(sx^{3}+\dots+ t)=0,\text{ and }\\(x+p/q)(ux^{4} +\dots +v)=0. \)

Multiplying both by q,

\(\displaystyle (qx+p)(sx^{3}+\dots+t)=0,\text{ and}\\(qx+p)(ux^{4}+\dots+v)=0.\)

Equating leading coefficients, qs = 75 so q = 1,3,5,25 or 75 and qu = 12 so q = 1,2,3,4,6 or 12.

q can't equal 1 since that would make k an integer, so q = 3.

Equating constants, pt = 12 so p = 1,2,3,4,6 or 12 and pv = 75 so p = 1,3,5,25 or 75.

p can't equal 3 since that would make k an integer, so p = 1, implying that k = -1/3.

Tiggsy Apr 27, 2019

#1**+3 **

Best Answer

Let k = -p/q, (where p and q are integers), then (x + p/q) will be a factor of both polynomials, so

\(\displaystyle (x+p/q)(sx^{3}+\dots+ t)=0,\text{ and }\\(x+p/q)(ux^{4} +\dots +v)=0. \)

Multiplying both by q,

\(\displaystyle (qx+p)(sx^{3}+\dots+t)=0,\text{ and}\\(qx+p)(ux^{4}+\dots+v)=0.\)

Equating leading coefficients, qs = 75 so q = 1,3,5,25 or 75 and qu = 12 so q = 1,2,3,4,6 or 12.

q can't equal 1 since that would make k an integer, so q = 3.

Equating constants, pt = 12 so p = 1,2,3,4,6 or 12 and pv = 75 so p = 1,3,5,25 or 75.

p can't equal 3 since that would make k an integer, so p = 1, implying that k = -1/3.

Tiggsy Apr 27, 2019