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JEE Main 202124 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Quadratic Equation

The number of the real roots of the equation (x + 1) ^ 2 + | x - 5...

The number of the real roots of the equation (x+1)2+∣x−5∣=274{\left(x + 1\right)}^{2} + | x - 5 | = \dfrac{27}{4} is ________.

Correct answer

0

Step-by-step solution

Case-II

x≤5x \le 5

(x+1)2−(x−5)=274{\left(x + 1\right)}^{2} - \left(x - 5\right) = \dfrac{27}{4}

(x+1)2−(x+1)−34=0{\left(x + 1\right)}^{2} - \left(x + 1\right) - \dfrac{3}{4} = 0

x+1=32,−12x + 1 = \dfrac{3}{2}, - \dfrac{1}{2}

x=12,−32x = \dfrac{1}{2}, - \dfrac{3}{2}

Case-IIII

x>5x > 5

\left(x + 1\right) + \left(x - 5\right) = \dfrac{27}{4}</p><p></p><p>{\left(x + 1\right)}^{2} + \left(x + 1\right) - \dfrac{51}{4} = 0</p><p></p><p>x = \dfrac{- 1 \pm \sqrt{52}}{2}(rejectedas(rejected asx > 5)

So, the equation have two real root.

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