Let a and b be the roots of the equation x^2-mx+2=0. Suppose that a+1/b and b+1/a are the roots of x^2-px+q=0. What is q?

From Vieta's formulas we know that when [tex]x_1[/tex] and [tex]x_2[/tex] are the roots of a quadratic equation:

[tex]Ax^2+Bx+C=0[/tex]

then:

[tex]x_1x_2=\dfrac{C}{A}[/tex]

In the first equation [tex]A=1[/tex], [tex]C=2[/tex], [tex]x_1=a[/tex] and [tex]x_2=b[/tex] so:

[tex]ab=\dfrac{2}{1}\\\\\\\boxed{ab=2}[/tex]

For the second one we have: [tex]A=1[/tex], [tex]C=q[/tex], [tex]x_1=a+\dfrac{1}{b}[/tex] and [tex]x_2=b+\dfrac{1}{a}[/tex] so:

[tex]x_1x_2=\dfrac{C}{A}\\\\\\x_1x_2=\dfrac{q}{1}\\\\\\\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{a}\right)=q\\\\\\ab+a\cdot\dfrac{1}{a}+\dfrac{1}{b}\cdot b+\dfrac{1}{b}\cdot\dfrac{1}{a}=q\\\\\\ab+\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{1}{ab}=q\\\\\\ab+1+1+\dfrac{1}{ab}=q\\\\\\2+1+1+\dfrac{1}{2}=q\\\\\\\boxed{q=4\dfrac{1}{2}=4.5}[/tex]