Key Concepts and Formulas

• Absolute Value Definition: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant: The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $D = b^2 - 4ac$. The nature of the roots depends on the discriminant.

Step-by-Step Solution

Step 1: Identify the Absolute Value and Define Cases

We are given the equation $(x+1)^2 + |x-5| = \frac{27}{4}$. The absolute value term $|x-5|$ changes its behavior at $x=5$. Therefore, we must consider two cases: $x \ge 5$ and $x < 5$.

Step 2: Case 1: $x \ge 5$

If $x \ge 5$, then $x-5 \ge 0$, so $|x-5| = x-5$.

Step 3: Substitute and Simplify the Equation for Case 1

Substitute $|x-5|$ with $(x-5)$ in the original equation: $(x+1)^2 + (x-5) = \frac{27}{4}$ Expand $(x+1)^2$: $x^2 + 2x + 1 + x - 5 = \frac{27}{4}$ Combine like terms: $x^2 + 3x - 4 = \frac{27}{4}$ Multiply both sides by 4 to eliminate the fraction: $4(x^2 + 3x - 4) = 27$ $4x^2 + 12x - 16 = 27$ $4x^2 + 12x - 43 = 0$

Step 4: Solve the Quadratic Equation for Case 1

Using the quadratic formula with $a=4$, $b=12$, and $c=-43$: $x = \frac{-12 \pm \sqrt{12^2 - 4(4)(-43)}}{2(4)}$ $x = \frac{-12 \pm \sqrt{144 + 688}}{8}$ $x = \frac{-12 \pm \sqrt{832}}{8}$ $x = \frac{-12 \pm \sqrt{64 \cdot 13}}{8}$ $x = \frac{-12 \pm 8\sqrt{13}}{8}$ $x = \frac{-3 \pm 2\sqrt{13}}{2}$ So, the two potential roots are $x_1 = \frac{-3 + 2\sqrt{13}}{2}$ and $x_2 = \frac{-3 - 2\sqrt{13}}{2}$.

Step 5: Check if the Roots Satisfy the Condition $x \ge 5$ for Case 1

Since $\sqrt{13} \approx 3.6$, $2\sqrt{13} \approx 7.2$. Thus, $x_1 = \frac{-3 + 2\sqrt{13}}{2} \approx \frac{-3 + 7.2}{2} \approx \frac{4.2}{2} \approx 2.1$ $x_2 = \frac{-3 - 2\sqrt{13}}{2} \approx \frac{-3 - 7.2}{2} \approx \frac{-10.2}{2} \approx -5.1$ Since $x_1 \approx 2.1$ and $x_2 \approx -5.1$, neither of these roots satisfies the condition $x \ge 5$. Therefore, there are no real roots in this case.

Step 6: Case 2: $x < 5$

If $x < 5$, then $x-5 < 0$, so $|x-5| = -(x-5) = 5-x$.

Step 7: Substitute and Simplify the Equation for Case 2

Substitute $|x-5|$ with $(5-x)$ in the original equation: $(x+1)^2 + (5-x) = \frac{27}{4}$ Expand $(x+1)^2$: $x^2 + 2x + 1 + 5 - x = \frac{27}{4}$ Combine like terms: $x^2 + x + 6 = \frac{27}{4}$ Multiply both sides by 4 to eliminate the fraction: $4(x^2 + x + 6) = 27$ $4x^2 + 4x + 24 = 27$ $4x^2 + 4x - 3 = 0$

Step 8: Solve the Quadratic Equation for Case 2

Using the quadratic formula with $a=4$, $b=4$, and $c=-3$: $x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)}$ $x = \frac{-4 \pm \sqrt{16 + 48}}{8}$ $x = \frac{-4 \pm \sqrt{64}}{8}$ $x = \frac{-4 \pm 8}{8}$ So, the two potential roots are $x_3 = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2}$ and $x_4 = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}$.

Step 9: Check if the Roots Satisfy the Condition $x < 5$ for Case 2

Since $x_3 = \frac{1}{2}$ and $x_4 = -\frac{3}{2}$, both of these roots satisfy the condition $x < 5$. Therefore, there are two real roots in this case.

Step 10: Combine the Results from Both Cases

Case 1 yielded no real roots. Case 2 yielded two real roots: $x = \frac{1}{2}$ and $x = -\frac{3}{2}$. Therefore, the total number of real roots is 2.

Common Mistakes & Tips

• Forgetting to Check Case Conditions: Always verify that the solutions obtained from each case satisfy the condition defined for that case.
• Sign Errors: Be careful with signs when expanding and simplifying equations, especially when dealing with the absolute value.
• Arithmetic Errors: Double-check calculations to avoid simple arithmetic errors, especially when applying the quadratic formula.

Summary

We solved the given equation by considering two cases based on the value of $x$ relative to 5. We found no real roots for $x \ge 5$, and two real roots ($x = 1/2$ and $x = -3/2$) for $x < 5$. Therefore, the equation has a total of 2 real roots.

The final answer is \boxed{2}.