Multiple choice

The number of quadratic equations which are unchanged by squaring their roots, is:

  1. $2$
  2. $4$
  3. $6$

  4. None of these
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B Correct answer
Explanation

For a quadratic equation to remain unchanged when its roots are squared, the set of roots {a, b} must be identical to the set of squared roots {a^2, b^2}. This leads to two possibilities: either each root is equal to its own square (giving roots {0,0}, {1,1}, or {0,1}), or the roots are mapped to each other's squares (giving the complex cube roots of unity {w, w^2}). This results in exactly 4 such quadratic equations.

AI explanation

Let the quadratic equation be x^2 + b x + c = 0 with roots r and s. The equation formed by squaring the roots is x^2 - (r^2 + s^2)x + r^2 s^2 = 0. For the equation to remain unchanged the coefficients must be proportional meaning b = -k(r^2 + s^2), c = k(r^2 s^2), b = -k(r + s) and c = k(rs). From c = k(rs) = k(r^2 s^2) we get rs = 1 or rs = 0. If rs = 1 then k=1 and b = -(r + s) = -(r^2 + s^2) = -((r + s)^2 - 2) which means r + s = 2 or r + s = -1. This yields the two equations x^2 - 2x + 1 = 0 and x^2 + x + 1 = 0. If rs = 0 we test roots r=0, s=-b; the squared equation becomes x^2 + b^2x = 0, requiring proportionality b^2/b = 1 so b=1 and c=0. If k=-1 for rs=0, we get b=-1 and c=0. Counting these distinct valid equations (x^2 - 2x + 1, x^2 + x + 1, x^2 + x, x^2 - x) gives a total of 4. The number of such equations is 4.

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