Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $3x^{2}-6x+5=0$ then the equation whose roots are $(\alpha+\beta)$ and $\dfrac{2}{(\alpha+\beta)}$ is

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

For 3x^2 - 6x + 5 = 0, sum of roots (alpha + beta) = -(-6)/3 = 2. Product of roots (alpha * beta) = 5/3. The new roots are (alpha + beta) = 2 and 2/(alpha + beta) = 2/2 = 1. The equation with roots 2 and 1 is x^2 - (2+1)x + (2*1) = 0, which is x^2 - 3x + 2 = 0.

AI explanation

Using Vieta's formulas, the sum of the roots alpha and beta for 3x^2 - 6x + 5 = 0 is alpha plus beta equals 6/3 equals 2. The new required roots are (alpha plus beta) and 2/(alpha plus beta), which evaluate to 2 and 2/2 equals 1. The sum of these new roots is 2 plus 1 equals 3, and their product is 2 multiplied by 1 equals 2. Forming the new quadratic equation using the formula x^2 minus (sum)x plus (product) equals 0 gives x^2 - 3x + 2 = 0.

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