Let α and β be the roots of 3x^2+4x+9=0. Then (1+α)(1+β) can be expressed in the form a/b, where a and b are coprime positive integers. Find a+b.
To find the value of (1+α)(1+β) in the given quadratic equation, we first need to find the values of α and β. Given the quadratic equation, 3x^2+4x+9=0, we can use the quadratic formula to find the roots:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values a = 3, b = 4, and c = 9 into the formula, we get:
x = (-4 ± √(4^2 - 4*3*9)) / (2*3)
= (-4 ± √(16 - 108)) / 6
= (-4 ± √(-92)) / 6
Since the discriminant is negative (Δ = b^2 - 4ac = 4^2 - 4*3*9 = 16 - 108 = -92 < 0), the roots of the quadratic equation are complex numbers. Let's simplify the expression of the roots:
x = (-4 ± √(-92)) / 6
= (-4 ± √(-1*92)) / 6
= (-4 ± √(-1)*√92) / 6
= (-4 ± 2i√23) / 6
= (-2/3) ± (1/3)i√23
Therefore, the roots of the quadratic equation are α = (-2/3) + (1/3)i√23 and β = (-2/3) - (1/3)i√23.
Now, let's find (1+α)(1+β):
(1+α)(1+β) = (1 + (-2/3) + (1/3)i√23) * (1 + (-2/3) - (1/3)i√23)
Using the distributive property, we can expand this expression:
(1+α)(1+β) = (1 - (2/3) + (1/3)i√23 - (2/3) - (1/3)i√23 + (2/3)(1/3)i^2*23)
Simplifying the expression further:
(1+α)(1+β) = 1 - (2/3) - (2/3) + (4/9)i^2*23
= 1 - (4/3) + 92/9
= 9/9 - 12/9 + 92/9
= 89/9
So, (1+α)(1+β) can be expressed as 89/9.
The required sum of the numerator and denominator is a+b = 89+9 = 98.