m & n are the roots of.....?

Question

ax^2+bx+c=0.Find the equation with the roots m^2+n^2 and
m^-2+n^-2 with the full method

Answer 1

I don't think DutchProf solved the problem that was given.
The stated problem begins by saying that m and n are the roots of ax^2+bx+c=0. Then it asks for the equation with two other roots that are expressed in terms of m and n.

If this is the correct interpretation of the problem, then one way to solve it is as follows (this is the method, not the solution):

Express m and n in terms of a, b, and c by applying the quadratic formula.

Substitute these expressions for m and n into the roots of the new equation (those roots being m^2+n^2 and m^-2+n^-2).

With these roots expressed in terms of a, b, and c, write the new equation by taking the product of x minus one root and x minus the other root (as DutchProf did, except that he used the roots expressed in terms of m and n, rather than a, b, and c) and setting it equal to 0.

Simplify the product determined in the preceding step.

If you can follow those steps, you should have the answer to the problem you stated in your question.

Answer 2

m and n being roots of ax^2 + bx + c means that:

m + n = -b/a
mn = c/a

m^2 + n^2 = (m + n)^2 - 2mn = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2
(m + n)/mn = 1/m + 1/n = -b/c
(1/m + 1/n)^2 = 1/m^2 + 2/mn + 1/n^2 = b^2/c^2
1/m^2 + 1/n^2 = b^2/c^2 - 2a/c = (b^2 - 2ac)/c^2

A quadratic equation with m^2 + n^2 and m^(-2) + n^(-2) as roots is then:

(x - {(b^2 - 2ac)/a^2})(x - {(b^2 - 2ac)/c^2}) = 0.

Alas, I am tired. I leave it to others more willing to simplify this quadratic equation into one containing no fractions.

Answer 3

As, m and n being roots of ax^2 + bx + c means that:

m + n = -b/a
mn = c/a

m^2 + n^2 = (m + n)^2 - 2mn = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2
1/m^2 + 1/n^2 = b^2/c^2 - 2a/c = (b^2 - 2ac)/c^2

A quadratic equation with m^2 + n^2 and m^(-2) + n^(-2) as roots is then:

(x - {(b^2 - 2ac)/a^2})(x - {(b^2 - 2ac)/c^2})

Answer 4

ax^2 + bx + c = 0

must be equivalent to

(x - m^2 - n^2) (x - m^(-2) - n^(-2)) = 0

so

x^2 - (m^2 + n^2) x - (m^(-2) + n^(-2)) x + (m^2 + n^2)(m^(-2) + n^(-2)) = 0

x^2 - (m^2 + n^2 + m^(-2) + n^(-2)) x + (2 + (m/n)^2 + (n/m)^2) = 0

So
a = 1
b = m^2 + n^2 + 1/m^2 + 1/n^2
c = 2 + (m/n)^2 + (n/m)^2.

Answer 5

PIE=BOX HALF TOO FOR PIECES 4 SALE LXY SG PIE RT 9 =GOD KNOWS WHAT/?

Answer 6

m2 and n2