Multiple choice

If $\alpha, \beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$

  1. $\alpha^{3}+\beta^{3}=\dfrac{3abc-b^{3}}{a^{3}}$
  2. $\alpha^{3}+\beta^{3}=\dfrac{b^{3}-3abc}{a^{3}}$
  3. $\alpha^{2}+\beta^{2}=\dfrac{2ac-b^{2}}{a^{2}}$
  4. $\dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}=\dfrac{b^{2}-2ac}{c^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
AI explanation

Using the sum and product of roots for ax^2 + bx + c = 0, we know alpha + beta = -b/a and alpha * beta = c/a. The sum of cubes formula states alpha^3 + beta^3 = (alpha + beta)^3 - 3*alpha*beta*(alpha + beta). Substituting the known values gives (-b/a)^3 - 3(c/a)(-b/a), which simplifies to -b^3/a^3 + 3abc/a^3. Combining these terms over a common denominator results in (3abc - b^3)/a^3.

Keep practicing — related questions