# how to write a quadratic if the product of the roots is -3?

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We can write a quadratic equation with roots a and b as:

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

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

as the product of the roots is -3.

=> x^2 - ( a + b) - 3 = 0

Now a and b can have several values which satisfy the condition that their product is -3, e.g. (-1 , 3) , ( 3/2 , -2) etc. We are not given any further information about what the roots are.

Therefore the quadratic equation is

**x^2 - (a + b) - 3 = 0**

Since we know that the quadratic equation has 2 roots and the product of the roots is -3, we'll apply Viete's relations, to identify the coefficients of the quadratic.

From the 2nd Viete's relation, we'll identify the coefficients c and a.

x1*x2 = c/a

x1*x2 = -3

-3 = c/a,

we'll put a = 1 => c = -3.

The sum is x1 + x2 = -b

x1*x2=-3

x2=-3/x1

x1 - 3/x1=-b

x1^2 + b*x1 - 3 = 0

delta = b^2 + 12

To create a perfect square, we'll put b^2 = 4 => b = 2

delta = 4 + 12 = 16

sqrt delta = 4

The quadratic equation is:

**x^2 - 2x -3= 0**