# Determine the quadratic equation, whose roots are -7 and 7?Determine the quadratic equation, whose roots are -7 and 7?

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### 5 Answers

You need to remember the factored form of a quadratic equation such that:

Since the problem provides the roots and yields:

**Hence, evaluating the quadratic equation under the given conditions yields .**

A quadratic equation with two roots can be expressed in terms of multiple of two factors as follows:

(x - x1)(x - x2) = 0

Where x1 and x2 are the two roots of the equation.

Substituting given values of the roots in the above equation:

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

x^2 - 7^2 = 0

x^2 - 49 = 0

Thus the required quadratic equation is:

x^2 - 49 = 0

The quadratic equation

(x-7)(x+7) = x*x +7x-7x -7*7=x^2-49

The quadratic equation with roots a and b are given by:

(x-a))x-b) = 0

x^2-(a+b)x + ab = 0.......(1)

a= -7 and b= 7

So substituting in (1) we get:

x^2-(-7+7)x +(-7(7) = 0

x^2 - 9 = 0 is the required quadratic equation.

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

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

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

We'll determine the second degree equation, when we know the sum and the product of it's roots:

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

So the equation will be:

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

**X^2 -49=0**

We also could determine the quadratic using the fact that a polynomial of n-th degree could be written as a product of "n" linear factors (there are as many factors as the degree of polynom is).

x1 and x2 are the roots of the equation of 2nd degree, so we'll write the equation as a product of 2 linear factors.

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

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

The product is a difference of squares:

**x^2-49=0**

The quadratic is **x^2-49.**