Multiple choice

If $\alpha ,\beta $ are the roots of the quadratic equation ${ x }^{ 2 }+px+q=0$, then the values of ${ \alpha }^{ 3 }+{ \beta }^{ 3 }$ and ${ \alpha }^{ 4 }+{ \alpha }^{ 2 }{ \beta }^{ 2 }+{ \beta }^{ 4 }$ are respectively

  1. $3pq-{ p }^{ 3 }$ and ${ p }^{ 4 }-3{ p }^{ 2 }q+3{ q }^{ 2 }$
  2. $-p\left( 3q-{ p }^{ 2 } \right) $ and $\left( { p }^{ 2 }-q \right) \left( { p }^{ 2 }+3q \right) $
  3. $pq-4$ and ${ p }^{ 4 }-{ q }^{ 4 }$
  4. $3pq-{ p }^{ 3 }$ and $\left( { p }^{ 2 }-q \right) \left( { p }^{ 2 }-3q \right) $
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D Correct answer
Explanation

alpha + beta = -p, alpha * beta = q. alpha^3 + beta^3 = (alpha + beta)^3 - 3*alpha*beta*(alpha + beta) = (-p)^3 - 3*q*(-p) = 3pq - p^3. alpha^4 + alpha^2*beta^2 + beta^4 = (alpha^2 + beta^2)^2 - alpha^2*beta^2 = ((-p)^2 - 2q)^2 - q^2 = (p^2 - 2q)^2 - q^2 = p^4 - 4p^2q + 4q^2 - q^2 = p^4 - 4p^2q + 3q^2 = (p^2 - q)(p^2 - 3q).

AI explanation

Using Vieta's formulas, the sum of the roots is (alpha + beta) = -p and the product is (alpha * beta) = q. We apply the sum of cubes formula to get alpha^3 + beta^3 = (alpha + beta)^3 - 3(alpha + beta)(alpha * beta), which substitutes to (-p)^3 - 3(-p)(q) = -p^3 + 3pq. Next, we evaluate alpha^4 + alpha^2 * beta^2 + beta^4 by rewriting it as (alpha^2 + beta^2)^2 - alpha^2 * beta^2, where (alpha^2 + beta^2) equals (alpha + beta)^2 - 2(alpha * beta) = p^2 - 2q. Substituting these into the squared expression gives (p^2 - 2q)^2 - q^2 = p^4 - 4p^2*q + 3q^2, which factors perfectly into (p^2 - q)(p^2 - 3q).

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