If a and b are the roots of the equation x^2-px+q=0, form the equation whose roots are α/β^2 and β/α^2?

Answer 1

Let α = a/b and β = b/a. Then, using the given roots a and b, we find the new roots:

α/β² = a/b / (b/a)² = a/b / (b²/a²) = a/b * a²/b² = a³/b³

β/α² = b/a / (a/b)² = b/a / (a²/b²) = b/a * b²/a² = b³/a³

Now, let's use Vieta's formulas to find the symmetric sums of the new roots. The new sum of roots (α/β² + β/α²) is:

a³/b³ + b³/a³ = a³/a³ + b³/b³ = (a³ + b³)/(a³b³)

By using the original equation and Vieta's formulas, we can rewrite p and q as following:

a + b = p
ab = q

We will now simplify the sum of the new roots (α/β² + β/α²) in terms of p and q:

a³ + b³ = (a + b)(a² - ab + b²) = p(a² - ab + b²)

ab(a² + b²) = q(p² - 2q)

Now, let's find the product of the new roots (α/β² * β/α²):

(a³/b³) * (b³/a³) = (a³b³)/(a³b³) = 1

Now, we have the sum and product of the new roots, so we can form the new equation:

x² - (sum of new roots)x + (product of new roots) = 0

Finally, we get the new equation:

x² - [ (a³ + b³) / (a³b³) ] x + 1 = 0

x² - [p(a² - ab + b²)/q]x + 1 = 0