The number of the real roots of the equation (x + 1)² + |x - 5| = 27/4 is ________.

Solution

Maths sir · 4 steps

Split at x = 5 to remove the modulus, solve each piece, then keep only roots that sit in their own piece.

1. Case x ≥ 5. Here (x+1)² ≥ 36, so the left side is at least 36, which is far more than 27/4 = 6.75. This case gives no root.
2. Case x < 5. Now |x - 5| = 5 - x, so the equation becomes x² + 2x + 1 + 5 - x = 27/4. Multiply by 4 and tidy up.

   4x² + 4x - 3 = 0 ⇒ (2x + 3)(2x - 1) = 0

3. Check the roots. This gives x = 1/2 and x = -3/2. Both are less than 5, so both are valid in this case.
4. So. The equation has exactly 2 real roots.

The common slip: 4. Solving the x ≥ 5 case and keeping its roots without checking them gives 4, not 2. Those roots are about 2.1 and -5.1, and neither is at least 5.

The idea to keep. With a modulus, break the line at the point where the inside is zero. A root found in a case counts only if it lies in that case's interval.

Answer from the NTA final answer key. Solution by PrepWiser.