Question

Consider the polynomials p(x) = 3x + 27x^2 and q(x)= 2 . Find the x -coordinate(s) of the point(s) of intersection of these two polynomials. What is the sum of these x -coordinates? (If there is only one point of intersection, give the corresponding x -coordinate.)

Answer 1

Answer:

The x -coordinate(s) of the point(s) of intersection of these two polynomials are $x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is $\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

Step-by-step explanation:

The intersections of the two polynomials, p(x) and q(x), are the roots of the equation p(x) = q(x).

Thus, $3x + 27x^2=2$ and we solve for x

$3x+27x^2-2=2-2\\27x^2+3x-2=0\\\left(27x^2-6x\right)+\left(9x-2\right)\\3x\left(9x-2\right)+\left(9x-2\right)\\\left(9x-2\right)\left(3x+1\right)=0$

Using Zero Factor Theorem: = 0 if and only if = 0 or = 0

$9x-2=0\\9x=2\\x=\frac{2}{9}$

$3x+1=0\\3x=-1\\x=-\frac{1}{3}$

The solutions are:

$x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is

$\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

We can check our work with the graph of the two polynomials.