Let p and q be real numbers such that p ≠ 0, p^3 ≠ q and p^3 ≠ − q. If α and β are nonzero complex numbers?

Question

satisfying α + β = − p and α^3 + β^3 = q, then a quadratic equation having α/β and β/α as its roots is?

Answer 1

α^3 + β^3 = q

⇒ (α + β)^3 − 3αβ (α + β) = q

⇒ − p^3 + 3pαβ = q

⇒ αβ = (q + p^3)/3p

⇒ x² - (α/β + β/α)x + (α/β * β/α) = 0

⇒ x² - ((α² + β²)/αβ)x + 1 = 0

⇒ x² - (((α + β)² −2αβ)/αβ)x + 1 = 0

⇒ x² - ((p² - 2((q + p^3)/3p))/((q + p^3)/3p))x + 1 = 0

⇒ (p^3 + q) x² − (3p^3 − 2p^3 − 2q) x + (p^3 + q) = 0

⇒ (p^3 + q) x² − (p^3 − 2q) x + (p^3 + q) = 0.



Hope this helps.!

Answer 2

Given

α + β = -p ....................... {1}

α^3 + β^3 = q .......................... {2}



From {2}

(α + β)(α^2 - αβ + β^2) = q

-p(α^2 - αβ + β^2) = q

α^2 - αβ + β^2 = -q/p ........................... {3}

α^2 + 2αβ + β^2 - 3αβ = -q/p

(α + β)^2 - 3αβ = q/p

(-p)^2 - 3αβ = q/p

p^2 - 3αβ = q/p

3αβ = p^2 - q/p

3αβ = (p^3 - q)/p

αβ = (p^3 - q)/(3p) ........................... {4}



Substitute {4} into {3}

α^2 + (p^3 - q)/p + β^2 = -q/p

α^2 + β^2 = -q/p - (p^3 - q)/p

α^2 + β^2 = (-p^3 - q + q)/p

α^2 + β^2 = (-p^3)/p

α^2 + β^2 = -p^2 ...................... {5}



If a quadratic equation having α/β and β/α as its roots, then

Sum of roots

= α/β + β/α

= (α^2 + β^2)/(αβ) ................... substitute {4} and {5}

= (-p^2)/[(p^3 - q)/(3p)]

= -p^3/[3(p^3 - q)]



Product of roots

= (α/β) (β/α)

= 1



Hence the quadratic equation is:

x^2 - (Sum of roots)x + (Product of roots) = 0

x^2 + x p^3/[3(p^3 - q)] + 1 = 0

3(p^3 - q) x^2 + x p^3 + 3(p^3 - q) = 0

Answer 3

product of roots = 1



sum = α/β + β/α = (α^2 + β^2) / (αβ)



now α + β = − p

α^3 + β^3 = q = (α + β)^3 - 3 (α + β) (αβ)



or q = -p^3 + 3p (αβ)



αβ = (q+p^3)/3p



(α^2 + β^2) = ( α + β)^2 - 2(αβ) = p^2 + 2((q+p^3)/3p = (2q+ 5p^3)/2p



so sum = (2q+ 5p^3)/2p/((q+p^3))/3p = 3(2q+ 5p^3)/2(p^3+ q)



so eqation



x^2- 3(2q+ 5p^3)/2(p^3+ q) x + 1 = 0



2x^2(q+p^3) + 3(2q+5p^3) + 2(p^3+ q) = 0

Answer 4

Four Possibilities that make sense: 1. It originated in British pubs as an abbreviation for "mind your pints and quarts." Supposedly this warned the barkeep to serve full measure, mark the customer's tab accurately, etc. 2. According to the Oxford English Dictionary, to be P and Q was a regional expression meaning top quality. It first shows up in a bit of doggerel from 1612: "Bring in a quart of Maligo, right true: And looke, you Rogue, that it be Pee and Kew." 3. The expression refers to the difficulty kids have distinguishing lower-case p and q, mirror images of each other. 4. Originated with printers who set headlines using movable type where the letters are mirror images of the regular alphabet. Lower-case P's look like Q's and vice versa. It was a reminder not to mix up the letters when putting them back in the rack after use. I think the last one is the most plausible.

Answer 5

The sum of your roots is a/b + b/a = (a^2 + b^2) / ab

The product of your roots is a/b * b/a = 1

Therefore, your quadratic is x^2 - [(a^2 + b^2) / ab] * x + 1 = 0

(ab)x^2 - (a^2 + b^2)x + (ab) = 0 is an equivalent quadratic



We need to find a^2 + b^2 and ab in terms of p and q. We have two equations to play with:

(a+b)^2 = p^2

a^2 + b^2 + 2ab = p^2 &&&&&&



a^3 + b^3 = q

(a+b)(a^2 - ab + b^2) = q

-p(a^2 - ab + b^2) = q

a^2 + b^2 - ab = -q/p

a^2 + b^2 - ab = -q/p &&&&&&



If you add those two equations I have marked &&&&&, you can find ab in terms of p and q.

Once you find ab, sub it in the second &&&& equation to find a^2 + b^2 in terms of p and q

Then, plug in what you got for a^2 + b^2 and ab into the quadratic.