Consider the quadratic equation ax^2+bx+c=0, where a, b, and c are rational numbers amd the quadratic has two distinct zeros . if one zero is rational, which of the following is true of the other zero. a. the other zero is rational b. the other zero is irrational c. the other zero is non -real

Let p and q be the two roots of ax^2+bx+c = 0. This means that (x-p) and (x-q) are the two factors

Rewrite ax^2+bx+c = 0 into x^2+(b/a)x+(c/a) = 0 which is the result of dividing everything by 'a', so the leading coefficient is 1.

Equate the two expressions and expand the left side.

(x-p)(x-q) = x^2+(b/a)x+(c/a)

x^2-px-qx+pq = x^2+(b/a)x+(c/a)

x^2+(-p-q)x+pq = x^2+(b/a)x+(c/a)

We have the two equations

-p-q = b/a and p*q = c/a

We can isolate q to get the following

-p-q = b/a ---> -q = (b/a)+p ---> q = (-b-pa)/a

p*q = c/a ---> q = c/(pa)

The equations shown above in bold result in q being rational if p is rational. This is due to the fact that doing any operation of addition, subtraction, multiplication or division of rational numbers leads to some other rational number. So if 'a' and 'c' are rational, then so is c/a. If p and 'a' are rational, then so is p*a. Etc etc.

In short, p being rational leads to q being rational.

Therefore, both roots are rational.

the image of (6, 9) under a dilation is (4, 6). the scale factor is

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: