Write a quadratic equation that has integer coefficients and has solutions 2 and -2.?

Question

THIS ONE HAS MY HEAD SPINNING, HELP PLEASE!
a. x^2 + 4 = 0

b. x^2 - 4x + 4 = 0

c. x^2 - 4 = 0

Answer 1

c

x^2 - 4 = 0
(x-2)(x+2)=0
x=2,-2

Answer 2

(X-2)(X+2) = 0 = x^2 +4

Answer 3

since the solutions of the equation are 2 and -2, then we can write x = 2 and x = -2 it implies that the factors of the equation are (x - 2) and (x + 2). multipliying the two factor
(x - 2)(x + 2) = x ^ 2 - 4
then the correct answer is x^2 - 4 = 4.

general principle is given the solutions of an equation we say that this are the zero's of the function ( that is they are the factors of the equation)
eg if the solutions of an equation are a, b, c, d, ... then the equation who have the solution is the equation

(x - a)(x - b)(x - c)(x - d) ... = 0

Answer 4

a quadratic equation that has solutions 2 and -2 is :

(x -2)(x+2) = x^2 -4 =0

Answer 5

(X - 2)(X + 2) = 0, which is the same as:
X^2 - 4 = 0

Answer 6

a. x^2 + 4 = 0
x^2=-4
your answers will be imaginary, so that's out

b. x^2 - 4x + 4 = 0
(x-2)^2=0
x-2=0
x=2
only one answer, so that's out

c. x^2 - 4 = 0
(x+2)(x-2)=0
x=-2 and x=2
perfect

Answer 7

(x-2)(x+2) = 0

x^2 -4 = 0

Answer 8

Stop that head spinning! It'll wring your neck.

Given that a, b, c .... are real, rational zeros of a polynomial, then the polynomial with smallest integer coefficients that has these zeros is

(x-a)(x-b)(x-c) etc multiplied by the least common denominator of a, b, c, ...

Answer 9

(x - 2)(x + 2) = 0
x² + 2x - 2x - 4 = 0
x² - 4 = 0
OPTION c