Let α and β be the roots of 3x2+4x+9=0. Then (1+α)(1+β) can be expressed in the form ab, where a and b are coprime positive integers. Find a+b.
since the roots are
-2/3 ±√23/3 i
1+α = 1/3 + √23/3 i
1+β = 1/3 - √23/3 i
(1+α)(1+β) = 1/9 + 23/9 = 24/9 = 8/3
To find the value of (1+α)(1+β), we can use the fact that the product of the roots of a quadratic equation can be found using the constant term and the coefficient of the linear term.
Given the quadratic equation 3x^2 + 4x + 9 = 0, we can find the product of the roots α and β by dividing the constant term (9) by the coefficient of the quadratic term (3). So, the product αβ is given by:
αβ = constant term / coefficient of quadratic term
αβ = 9 / 3
αβ = 3
Now, we need to find the value of (1+α)(1+β). Expanding this expression, we get:
(1+α)(1+β) = 1 + α + β + αβ
Substituting the value of αβ that we found earlier, we get:
(1+α)(1+β) = 1 + α + β + 3
Since α and β are the roots of the quadratic equation 3x^2 + 4x + 9 = 0, we can use Vieta's formulas to find the sum of the roots. Vieta's formulas state that for a quadratic equation ax^2 + bx + c = 0, the sum of the roots is given by -b/a.
In our case, the sum α + β is equal to:
α + β = -b/a
α + β = -4/3
Substituting this value into (1+α)(1+β), we get:
(1+α)(1+β) = 1 + α + β + 3
(1+α)(1+β) = 1 + (-4/3) + 3
(1+α)(1+β) = 4/3 + 1 + 3
(1+α)(1+β) = 4/3 + 3/3 + 9/3
(1+α)(1+β) = 16/3
So, (1+α)(1+β) can be expressed as 16/3. To express it in the form ab, we can write it as 16/3 = 16 * (1/3). Therefore, a = 16 and b = 3.
To find a + b, we simply add the values of a and b:
a + b = 16 + 3
a + b = 19
Therefore, the value of a + b is 19.