There's not a simple way to factorize \(2x^2 - 3x - 1 = 0\), so instead we simplify \(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{(a+b)^2-2ab}{ab}=\frac{(a+b)^2}{ab}-2\)
Using Vieta's Formula, a+b = 3/2 and ab = -1/2.
Plug these in to get \(\frac{(3/2)^2}{-1/2}-2=\frac{9/4}{-1/2}-2=-\frac{9}{2}-2=\frac{-13}{2}\)
.You could plug in the values of 2 and 3 to create a system of equations with two variables.
However, since this problem is only asking for the value of m, we can use Vieta's Formula instead.
Vieta's Formula states that the product of the roots in the equation \(ax^2+bx+c=0\) equals \(\frac{c}{a}\). In our equation, c = 2m and a = 3.
So \(\frac{2m}{3}=2*3\).
Solve to get m = 9.