Multiple choice

If roots of the equations $ax^2- 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real, then

  1. $ac = b^2$
  2. $4b^2 - ac = 0$
  3. $a = b, c = 0$
  4. $a = b = 0$
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A Correct answer
Explanation

For the roots to be real, the discriminant must be >= 0. For ax^2 - 2bx + c = 0, D = 4b^2 - 4ac >= 0 => b^2 >= ac. For bx^2 - 2sqrt(ac)x + b = 0, D = 4ac - 4b^2 >= 0 => ac >= b^2. Both conditions imply ac = b^2.

AI explanation

For the second equation bx squared minus two times the square root of ac times x plus b equals zero to have real roots, its discriminant must be greater than or equal to zero. The discriminant is calculated as four times ac minus b squared, leading to the condition four ac minus b squared is greater than or equal to zero. This implies that ac equals b squared.

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