Determine the second grade equation, when knowing that x1=-7 and x2=7. The determination must be done in two ways, at least.

A second degree equation is also known as a quadratic equation. A second dgree or quadratic equation is of the type:ax^2+bx+c=0

If the roots of the equation x1 and x2 are known, then we can determine the the quadratic equation.

If x1 and x2 are the roots, then it satisfies the equation.

To determine the equation:

First method : The relations between the roots x1 abd x2 and the coefficients of x^2 , x and free term in  the quadratic equation ax^2+bx+c=0  are:

x1+x2= -b/a ...........(1)and

x1*x2 = c/a.............(2)

Given x1=-7 and x2=7. Substitutingthe values  in the above equations, we get:

x1+x2=-7+7=0=-b/a. Therefore, b=0

-7*7=c/a or c=-49a

Therefore,  the equation is ax^2-49a=0 or dividing by a we get:

x^2-49=0 which is the required second degree equation.

2nd method:

Let f(x) = 0 be the required equation.

If x1 is the zero of the function f(x) , then x-x1 is a factor of f(x). Similarly if x2 is a zero of the the function f(x), then (x-x2) is factor of  f(x).

So f(x) = k(x-x1)(x-x2) if x1 and x2 are the zeros of the function f(x). Therefore, the required equation of 2nd degree is :

f(x)=0 takes the form k(x-x1)(x-x2)=0

Dividind by k, we get: (x-x1)(x-x2)=0..................(3)

Substitute x1=-7 and x2=+7 in the above equation (3), we get:

(x+7)(x-7)=0 or

x^2-7x+7x-49=0 or

x^2-49 =0 is required 2nd degree equation.