Multiple choice

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

  1. $x^2 \, - \, 3x \, + \, 11 \, = \, 0 $
  2. $x^2 \, +\, 6x \, + \, 11 \, = \, 0 $
  3. $x^2 \, - \, 6x \, + \, 11 \, = \, 0 $
  4. $x^2 \, + \, 3x \, + \, 11 \, = \, 0 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the new roots be y = x + 2, so x = y - 2. Substitute this into the original equation: (y-2)^2 - 2(y-2) + 3 = 0. Expanding gives y^2 - 4y + 4 - 2y + 4 + 3 = 0, which simplifies to y^2 - 6y + 11 = 0.

AI explanation

For the equation x^2 - 2x + 3 = 0, the sum of the roots alpha + beta is 2 and the product alpha * beta is 3. The new roots are each increased by 2, so their sum is alpha + 2 + beta + 2 = (alpha + beta) + 4 = 2 + 4 = 6. The new product is (alpha + 2)(beta + 2) = alpha * beta + 2(alpha + beta) + 4 = 3 + 4 + 4 = 11. Forming the quadratic equation with this sum and product gives x^2 - 6x + 11 = 0.

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