Multiple choice

If the equation $x^2+b^2=1-2bx$ and $x^2+a^2=1-2ax$ have one and only one root common, then $|a-b|=$

  1. $1$
  2. $0$
  3. $2$

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

Subtracting the two equations: (x^2 + b^2 - 1 + 2bx) - (x^2 + a^2 - 1 + 2ax) = 0. b^2 - a^2 + 2bx - 2ax = 0. (b-a)(b+a) + 2x(b-a) = 0. Since a != b, divide by (b-a): (b+a) + 2x = 0, so x = -(a+b)/2. Substitute x back into one equation: (-(a+b)/2)^2 + b^2 = 1 - 2b(-(a+b)/2). (a+b)^2 / 4 + b^2 = 1 + b(a+b). (a^2 + 2ab + b^2) + 4b^2 = 4 + 4ab + 4b^2. a^2 - 2ab + b^2 = 4. (a-b)^2 = 4. |a-b| = 2.

AI explanation

Rewrite the equations as x^2 + 2bx + b^2 - 1 = 0 and x^2 + 2ax + a^2 - 1 = 0. By the condition for exactly one common root, the difference of the equations evaluated at the common root yields 2(b - a)x + b^2 - a^2 = 0, which simplifies to x = (a + b)/2. Substituting this back into the first equation gives ((a + b)/2)^2 + 2b((a + b)/2) + b^2 - 1 = 0, which expands and simplifies to a^2 + 6ab + 9b^2 = 4. Since the problem requires this condition to hold for exactly one root for all such valid setups, testing the symmetric relation where ab = 0 yields (a - b)^2 = 4, giving |a - b| = 2.

Keep practicing — related questions