Multiple choice

If $\displaystyle \alpha \neq \beta$ but $\alpha ^{2}= 5\alpha -3, \beta ^{2}= 5\beta -3$, then the equation whose roots are $\displaystyle \frac{\alpha}{\beta} $ and $\dfrac{\beta}{\alpha} $ is

  1. $\displaystyle 3x^{2}-19x-3= 0.$
  2. $\displaystyle 3x^{2}+19x+3= 0.$
  3. $\displaystyle 3x^{2}-19x+3= 0.$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given alpha^2 - 5alpha + 3 = 0 and beta^2 - 5beta + 3 = 0, alpha and beta are roots of x^2 - 5x + 3 = 0. Sum of roots = 5, product = 3. We need the equation with roots alpha/beta and beta/alpha. Sum = (alpha^2 + beta^2) / (alpha*beta) = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta) = (25 - 6) / 3 = 19/3. Product = 1. Equation: x^2 - (19/3)x + 1 = 0, or 3x^2 - 19x + 3 = 0.

AI explanation

The given relations alpha^2 = 5*alpha - 3 and beta^2 = 5*beta - 3 mean alpha and beta are roots of x^2 - 5x + 3 = 0. By Vieta's formulas, the sum of the roots is alpha + beta = 5 and the product is alpha*beta = 3. The sum of the new roots is alpha/beta + beta/alpha = (alpha^2 + beta^2)/(alpha*beta) = ((alpha+beta)^2 - 2*alpha*beta)/(alpha*beta) = (25 - 6)/3 = 19/3. The product of the new roots is (alpha/beta)*(beta/alpha) = 1. The required quadratic equation is x^2 - (19/3)x + 1 = 0, which simplifies to 3x^2 - 19x + 3 = 0.

Keep practicing — related questions