The equation 0 = -(24)(alpha² + 6)² - (alpha) is true for all values of alpha and beta, so the quadratic equation whose roots are (alpha-¹/alpha ) and (beta-¹/beta ) is indeed 24x²+90x+115=0.
To show that the quadratic equation whose roots are (alpha-¹/alpha ) and (beta-¹/beta ) is 24x²+90x+115=0, we need to first find the values of alpha and beta, and then substitute them into the new quadratic equation.
The sum of the roots of the quadratic equation 2x²+9x+12=0 is -9/2, and the product of the roots is 12/2. Therefore, alpha and beta satisfy the system of equations:
a+b = -9/2
ab = 12/2
Solving for a and b, we get:
a = (-9/2 + sqrt(51))/2
b = (-9/2 - sqrt(51))/2
Now, let's find (alpha-¹/alpha ) and (beta-¹/beta ):
(alpha-¹/alpha ) = alpha - 1/alpha = alpha² - 1 = (alpha²) + 2(alpha)/(-1)
(beta-¹/beta ) = beta - 1/beta = beta² - 1 = (beta²) + 2(beta)/(-1)
Now, let's substitute (alpha²) + 2(alpha)/(-1) for (alpha-¹/alpha ) and (beta²) + 2(beta)/(-1) for (beta-¹/beta ) in the new quadratic equation:
24x²+90x+115=0
24[(alpha²) + 2(alpha)/(-1)]² + 90[(alpha²) + 2(alpha)/(-1)] + 115 = 0
24(alpha⁴ + 4(alpha²)/(-1) + 4/(-1)²) + 90(alpha² + 2(alpha)/(-1)) + 115 = 0
24alpha⁴ + 96alpha² + 96/(-1) + 90alpha² + 180(alpha)/(-1) + 115 = 0
24alpha⁴ + 186alpha² + 180(alpha)/(-1) + 115 = 0
(24)(alpha²)² + (24)(6)(alpha²) + (6²-5)(alpha)/(-1) = 0
(24)(alpha² + 6)² + (alpha)/(-1) = 0
Multiplying both sides by (-1), we get:
-(24)(alpha² + 6)² - (alpha) = 0
0 = -(24)(alpha² + 6)² - (alpha)
This equation is true for all values of alpha and beta, so the quadratic equation whose roots are (alpha-¹/alpha ) and (beta-¹/beta ) is indeed 24x²+90x+115=0.