Let a and b be the roots of the equation x^2 - mx + 2 = 0. Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0. What is q?
Let a and b be the roots of the equation x^2 - mx + 2 = 0. Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0. What is q?
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\(x^2 - mx + 2 = 0\)
p q
\(x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}\\ a=\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2}\\ b=\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}\)
\({\color{blue}x^2 - px + q =}\{x-(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2}+\frac{1}{\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}}\\ +(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2})(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}))\}\)
\(\times\{x-(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}+\frac{1}{\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2} }+(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2})(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}))\}\)
\(\sqrt{\frac{m^2}{4}-2}=z\)
\({\color{blue}x^2 - px + q =}\{x-(\frac{m}{2}+ z+\frac{1}{\frac{m}{2}- z} +(\frac{m}{2}+ z )(\frac{m}{2}- z ))\}\)
\(\times\{x-(\frac{m}{2}-z +\frac{1}{\frac{m}{2}+ z }+(\frac{m}{2}+ z )\times (\frac{m}{2}-z ))\}\)
\({\color{blue}x^2 - px + q =} (x-(a + 1/b + ab))(x-( b + 1/a + ab))\)
!
Let a and b be the roots of the equation x^2 - mx + 2 = 0.
Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0.
What is q?
Let a and b be the roots of the equation x^2 - mx + 2 = 0.
a+b=m
ab=2
Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0
(a + 1/b + ab) + (b + 1/a + ab) = p and (a + 1/b + ab) * (b + 1/a + ab) = q
(a + 1/b + ab) = (ab+1)/b + ab = (2+1)/b + 2 = 3/b + 2
likewise
(b + 1/a + ab) = 3/a + 2
so
\(q= (\frac{3}{a}+2) (\frac{3}{b}+2)\\ abq= (3+2a) (3+2b)\\ 2q=9+6b+6a+4ab\\ 2q=9+6(a+b)+4ab\\ 2q=9+6(m)+8\\ 2q=17+6m\\ q=3m+8.5\)
Maybe you can get rid of the m if you go further with the working.
But the question does not state whether you must get q as a number or as a function of m.