Which equation has roots of -3 and 5

Question

Which equation has roots of -3 and 5

 

Answer 1

Sherry,

  This is like 'Algebra Jeopardy'....Given the answer, what is the question!?

For a quadratic equation like x^2 + b*x + c =0 you end up with factoring usually like this:

(x+w)(x+z)=0

The the roots are x = -w, -z

The problem gives the roots  -w and -z so now just subsitute:

-w = -3  and -z = 5

The factored equation is:

(x+3)(x-5) = 0  or expanded it is:

X^2 +3*x -5*x -15 = 0

 

X^2 -2*x -15 = 0

Note: I have assumed a quadratic equation but there are many equaitons of other types with these roots.    For example:

x+3 =0    so x=-3

x-5 = 0  so x=5

(x+3)(x+3)(x-5) = 0  This is a third order polynomial with a double root:  x=-3

And so it goes...

BruceS

Answer 2

If an equation has only these two roots -3 and 5, you can assume that this is a quadratic equation ax^2 + bx + c = 0 where the first coefficient equals 1 (a=1). This equation can be factored in the following way: (x-x1)(x-x2) = 0 where x1, x2 are roots of the equation. Substituting the numbers into the equation you can easily find (x+3)(x-5) = 0 If your equation, except for -3 and 5, has other roots, you can assume that this a polynomial equation of higher degree. That polynomial can be factorized the same way if the roots are whole numbers.

Answer 3

if the roots are r1 and r2, then the simplest equation is the quadratic: (x-r1)(x-r2) = 0 since yours are -3 and 5, the simplest (quadratic) equation is: (x - -3)(x - 5) = 0 (x+3)(x-5) = 0 which expands to x²-2x-15 = 0

Answer 4

If the roots are -3 and 5

Then the formula is

(x-3)(x+5)=0

or

x²+2x-15=0