Multiple choice

Solve the following equations. $x^4\,-\,3x^2\,+\,2\,=\,0$, roots are

  1. $x\,=\,\pm\,\sqrt3,\,\pm\,1$
  2. $x\,=\,\pm\,\sqrt2,\,\pm\,1$
  3. $x\,=\,\pm\,\sqrt2,\,\pm\,3$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x^2. The equation becomes y^2 - 3y + 2 = 0, which factors to (y - 1)(y - 2) = 0. So y = 1 or y = 2. Since y = x^2, x^2 = 1 (x = +/- 1) or x^2 = 2 (x = +/- sqrt(2)).

AI explanation

Let y = x^2, which changes the equation into the quadratic form y^2 - 3y + 2 = 0. Factoring this gives (y - 2)(y - 1) = 0, meaning y = 2 or y = 1. Replacing y back with x^2 gives x^2 = 2 or x^2 = 1, so the roots are x = plus or minus the square root of 2 and x = plus or minus 1.

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