Giving out brainliest !!
Let k be a real constant. Note that (x, y) = (0,0) is always a solution to the system-

x+3y = kx
3x+y = ky

Find all values of k for which (0,0) is not the only solution.

Step-by-step explanation:

Trivial Solutions

We are dealing with a type of systems of equations with the following structure:

There is a trivial solution where x=y=0, but the interest of the problem is to find the conditions for a given system to have other solutions than the trivial. The idea is to transform the compatible determinate system to a compatible indeterminate system that accepts infinitely many solutions. This can be achieved by computing the determinant of the system

If this determinant is zero, the system is compatible indeterminate.

Let's analyze the given system:

Rearranging

Computing the determinant and equating to 0

Expanding the determinant

(1-k)^2-9=0

Rearranging

(1-k)^2=9

Taking the square root (with both possible signs)

1-k=\pm 3

Solving for k

k=1\pm 3

The two possible values of k are

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1