To solve the equation \( (x+1)^2 + |x-5| = \frac{27}{4} \), we will analyze it systematically.
- Break down the equation into its components: \( (x+1)^2 \) and \( |x-5| \).
- The equation involves an absolute value, which requires considering two cases for \( |x-5| \):
- Case 1: \( x \geq 5 \) implies \( |x-5| = x-5 \).
- Case 2: \( x < 5 \) implies \( |x-5| = 5-x \).
- For Case 1 (\( x \geq 5 \)), substitute \( |x-5| = x-5 \):
- \( (x+1)^2 + (x-5) = \frac{27}{4} \)
- Simplifying, we have:
- \( x^2 + 2x + 1 + x - 5 = \frac{27}{4} \)
- \( x^2 + 3x - 4 = \frac{27}{4} \)
- Clear the fraction by multiplying by 4:
- \( 4x^2 + 12x - 16 = 27 \)
- \( 4x^2 + 12x - 43 = 0 \)
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 4, b = 12, c = -43 \):
- \( x = \frac{-12 \pm \sqrt{12^2-4 \times 4 \times (-43)}}{2 \times 4} \)
- Calculate the discriminant:
- \( 144 + 688 = 832 \)
- \( x = \frac{-12 \pm \sqrt{832}}{8} \)
- The roots, calculated, are irrelevant as they fall outside the range \( x \geq 5 \).
- For Case 2 (\( x < 5 \)), substitute \( |x-5| = 5-x \):
- \( (x+1)^2 + (5-x) = \frac{27}{4} \)
- \( x^2 + 2x + 1 + 5 - x = \frac{27}{4} \)
- \( x^2 + x + 6 = \frac{27}{4} \)
- Clear the fraction by multiplying by 4:
- \( 4x^2 + 4x + 24 = 27 \)
- \( 4x^2 + 4x - 3 = 0 \)
- Again, apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 4, b = 4, c = -3 \):
- \( x = \frac{-4 \pm \sqrt{16 + 48}}{8} \)
- \( x = \frac{-4 \pm \sqrt{64}}{8} \)
- \( x = \frac{-4 \pm 8}{8} \)
- This gives:
- \( x = \frac{4}{8} = \frac{1}{2} \)
- \( x = \frac{-12}{8} = -\frac{3}{2} \)
- Both roots, \( \frac{1}{2} \) and \(-\frac{3}{2}\), satisfy \( x < 5 \).
Therefore, the number of real roots for the given equation is two, which falls within the expected range of 2,2.