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
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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
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