Multiple choice

If $\alpha \, and \, \beta$ are the roots of the equation $x^{2} \, - \, 2x \, + \, 3 \,= \, 0$ Find the equation whose roots are $\dfrac{\alpha \, - \,1}{\alpha \, + \,1 }\, , \dfrac{\beta \, - 1}{\beta \, + 1}$

  1.  $3x^{2} \, - \, 2x \, + \, 1 \,= \, 0$
  2.  $x^{2} \, - \, x \, + \, 3 \,= \, 0$
  3.  $5x^{2} \, - \, 2x \, + \, 3 \,= \, 0$
  4.  $x^{2} \, - \, 2x \, + \, 7 \,= \, 0$
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A Correct answer
Explanation

Let y = (x - 1)/(x + 1). Solving for x gives x = (1 + y)/(1 - y). Substituting this into the original equation (x^2 - 2x + 3 = 0) results in ((1 + y)/(1 - y))^2 - 2((1 + y)/(1 - y)) + 3 = 0. Simplifying this leads to 3y^2 - 2y + 1 = 0.

AI explanation

From the equation x^2 - 2x + 3 = 0, the sum of the roots is 2 and the product is 3. If a root of the new equation is y = (x - 1)/(x + 1), we can rearrange for x to get x = (1 + y)/(1 - y). Substituting this expression into the original equation gives ((1 + y)/(1 - y))^2 - 2((1 + y)/(1 - y)) + 3 = 0. Multiplying by (1 - y)^2 and simplifying yields 3y^2 - 2y + 1 = 0.

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