Question:
The number of the real roots of the equation \( (x+1)^2 + |x-5| = \dfrac{27}{4} \) is _______
The number of the real roots of the equation \( (x+1)^2 + |x-5| = \dfrac{27}{4} \) is _______
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Always split absolute value $|x-a|$ into two cases: $x \geq a$ and $x<a$ to solve the resulting equations separately.
Updated On: Apr 16, 2026
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Correct Answer: 2
Solution and Explanation
Case 1 ($x \geq 5$): $x^2 + 2x + 1 + x - 5 = 6.75 \Rightarrow x^2 + 3x - 10.75 = 0$. Roots are $\approx 2.1$ and $-5.1$. Neither is $\geq 5$.
Case 2 ($x<5$): $x^2 + 2x + 1 - x + 5 = 6.75 \Rightarrow x^2 + x - 0.75 = 0$.
Step 1: $4x^2 + 4x - 3 = 0 \Rightarrow (2x+3)(2x-1) = 0$.
Step 2: $x = 0.5$ or $x = -1.5$. Both are $< 5$. Total real roots = 2.
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