Multiple choice

If $c\neq 0$ and $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx +c=0$, then value of $\displaystyle \frac{a\alpha^2 + c}{a\alpha +b}+\frac{a\beta^2+ c}{a\beta+b}$ is

  1. $\displaystyle \frac{b(b^2-2ac)}{4a}$
  2. $\displaystyle \frac{(b^2-2ac)}{2a}$
  3. $\displaystyle \frac{b(b^2-2ac)}{a^2c}$
  4. $\displaystyle \frac{b(b^2-2ac)}{ac}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given ax^2 + bx + c = 0, roots alpha, beta satisfy a*alpha^2 + b*alpha + c = 0, so a*alpha^2 + c = -b*alpha. The expression becomes (-b*alpha)/(a*alpha + b) + (-b*beta)/(a*beta + b). This simplifies using symmetric sums of roots.

AI explanation

Using the root alpha in the original equation, a*alpha^2 + b*alpha + c = 0, we can rewrite the first numerator as a*alpha^2 + c = -b*alpha. This simplifies the first fraction to (-b*alpha)/(a*alpha + b). Applying the same logic to the second fraction for root beta gives (-b*beta)/(a*beta + b). Adding the fractions and using the Vieta's formulas sum (alpha + beta) = -b/a and product alpha*beta = c/a leads to the combined expression b(b^2 - 2ac)/(a^2*c).

Keep practicing — related questions