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

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Based on fact that a polynomial of n-th degree could be written as a product of "n" linear factors (they are as many factors as the degree of polynom is) and knowing that x1 and x2 are the roots of the equation of 2nd degree, than we could write the equation as a product of 2 linear factors, decomposing the second degree equation in this way:

a(x-x1)(x-x2)=0, where x1=7 and x2=-7

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

As we've seen, we deal with a difference of squares:

**x^2-49=0**

So the associated function is **f(x)=x^2-49.**

**The second method:Viete mathematical relations:**

x1+x2=S=-b/a=7+(-7)=0

x1*x2=P=c/a=7*(-7)=-49

where a,b,c, are the coefficients of the function f(x)=a*x^2 + b*x + c

We know to determine the second degree equation, when we know the sum and the product:

X^2 - S*X + P=0, butthe sum S=0 and the product P=-49

So the equation will be:

X^2 - 0*X +(-49)=0

**X^2 -49=0**

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.