Multiple choice

Find the roots of the following quadratic equations by using the quadratic formula $(x^2 - 2x)^2 - 4(x^2 - 2x) + 3= 0$

  1. $-1, 3, 1\pm \sqrt{2}$
  2. $1, 3, -1\pm \sqrt{2}$
  3. $1, 3, 1\pm \sqrt{2}$

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

Let y = x^2 - 2x. Equation: y^2 - 4y + 3 = 0 -> (y-3)(y-1) = 0. Case 1: x^2 - 2x = 3 -> x^2 - 2x - 3 = 0 -> (x-3)(x+1) = 0 -> x = 3, -1. Case 2: x^2 - 2x = 1 -> x^2 - 2x - 1 = 0 -> x = (2 +/- sqrt(4 + 4)) / 2 = 1 +/- sqrt(2).

AI explanation

Let y equal x squared minus 2x, which changes the equation to y squared minus 4y plus 3 equals 0. Factoring this gives y equal to 1 or y equal to 3, so x squared minus 2x equals 1 and x squared minus 2x equals 3. Solving x squared minus 2x minus 1 equals 0 with the quadratic formula gives x equal to 1 plus or minus the square root of 2, and solving x squared minus 2x minus 3 equals 0 gives x equal to 3 or x equal to negative 1. The complete set of roots is negative 1, 3, and 1 plus or minus the square root of 2.

Keep practicing — related questions