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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

The number of the real roots of the equation \((x+1)^{2}+|x-5|=\frac{27}{4}\) is ....... .

  1. A \(6\)
  2. B \(0\)
  3. C \(4\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

Case\(-I\) \(x \leq 5\) \((x+1)^{2}-(x-5)=\frac{27}{4}\) \((x+1)^{2}-(x+1)-\frac{3}{4}=0\) \(x+1=\frac{3}{2},-\frac{1}{2}\) \(x=\frac{1}{2},-\frac{3}{2}\) Case\(-II\) \(x >5\) \((x+1)+(x-5)=\frac{27}{4}\) \((x+1)^{2}+(x+1)-\frac{51}{4}=0\) \(x=\frac{-1 \pm \sqrt{52}}{2}(\)…
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