Multiple choice

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

  1. $2x^{2}+3x+2=0$
  2. $3x^{2}+2x+3=0$
  3. $2x^{2}-3x+2=0$
  4. $3x^{2}-2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha^2 - 2alpha + 3 = 0 and beta^2 - 2beta + 3 = 0, alpha and beta are roots of x^2 - 2x + 3 = 0. Thus, alpha + beta = 2 and alpha*beta = 3. The new roots are alpha/beta + beta/alpha = (alpha^2 + beta^2)/(alpha*beta) = ((alpha+beta)^2 - 2alpha*beta)/(alpha*beta) = (4 - 6)/3 = -2/3. The product is 1. The equation is x^2 - (sum)x + product = 0, which is x^2 + 2/3x + 1 = 0, or 3x^2 + 2x + 3 = 0.

AI explanation

Since alpha and beta satisfy x^2 = 2x - 3, they are the roots of the equation x^2 - 2x + 3 = 0, giving the sums alpha+beta=2 and alpha*beta=3. The required equation has roots alpha/beta and beta/alpha, so the sum of the new roots is (alpha^2 + beta^2)/(alpha*beta), which evaluates to [(alpha+beta)^2 - 2(alpha*beta)] / 3 to give 4/3 - 2 = -2/3. The product of the new roots is 1. Using the formula for forming a quadratic equation, x^2 - (sum)x + (product) = 0, we get x^2 + (2/3)x + 1 = 0, which scales to the final result of 3x^2 + 2x + 3 = 0.

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