IF $3{p}^{2}=5p+2$ and $3{q}^{2}=5q+2$ where $p\ne q$, then the equation whose roots are $3p-2q$ and $3q-2p$ is
- $3{x}^{2}-5x-100=0$
- $5{X}^{2}+3x+100=0$
- $3{x}^{2}-5x+100=0$
- $5{x}^{2}-3x-100=0$
Given 3p^2 - 5p - 2 = 0 and 3q^2 - 5q - 2 = 0, p and q are roots of 3x^2 - 5x - 2 = 0. Thus p+q = 5/3 and pq = -2/3. We need an equation with roots alpha = 3p-2q and beta = 3q-2p. Sum = alpha + beta = p+q = 5/3. Product = (3p-2q)(3q-2p) = 9pq - 6p^2 - 6q^2 + 4pq = 13pq - 6(p+q)^2 + 12pq = 25pq - 6(p+q)^2 = 25(-2/3) - 6(25/9) = -50/3 - 50/3 = -100/3. Equation: x^2 - (5/3)x - 100/3 = 0, or 3x^2 - 5x - 100 = 0.
Since p and q are distinct roots of 3x squared minus 5x minus 2 equals 0, their sum is 5/3 and their product is -2/3. The sum of the new roots, which are 3p minus 2q and 3q minus 2p, equals p plus q, resulting in 5/3. The product of the new roots simplifies to 13 times p times q minus 6 times the square of the sum of p and q, yielding 13 times -2/3 minus 6 times 25/9, which equals -100/3. Using the sum and product to form the new quadratic gives 3x squared minus 5x minus 100 equals 0. The result is 3x^2 - 5x - 100 = 0.