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FInd the quadratic equation each of whose roots is the sum of a root
and its reciprocal of the quadratic equation 2x^2 + 3x + 4 = 0.
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This problem is to apply the Vieta's theorem several times.
Let " a " and " b " be the roots of the given equation.
Then according to Vieta's theorem, the sum (a+b) is equal to : a + b = .
According to the same theorem, the product of the routs ab is equal to = 2.
OK.
HENCE, the projected equation has the value as one of its roots.
The other root of the projected equation is the sum of reciprocals = .
In this expression, we can replace a+b by and can replace ab by 2, since we just know it.
Making this replacement, you get = = .
So, the roots of the projected equation are and .
Hence, the sought equation is
= 0, or, EQUIVALENTLY, = 0.
If you want to have the sought equation with integer coefficients, you can multiply the last equation by 8,
leaving it equivalent.
In this way, you get the ANSWER:
the sought equation is (2x+3)*(4x+3) = 0 with the roots of and .
Solved.
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There is another way to solve the problem.
It is to find the roots of the given equation explicitly and then calculate their sum and the sum of reciprocals.
It is possible, but it will lead you through the forest of calculations with radicals and fractions of complex numbers with radicals.
So, this way is possible, but it is not elegant.
The way which I showed you in my post, is TRULY ELEGANT, and IT IS the ONLY expected way
as the problem should be solved and presented.
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May I ask you ?
This problem is advanced: it is destined for advanced students.
Advanced students are THOSE who love Math and love solving tricky problems on their own.
If so, then what is the reason for you, for an advanced student, post this problem
to the forum instead of solving it on YOUR OWN ?