Verified Solution • 2025–26 Session moderate

QJEEM21Ma18fc: JEE Main 2021 Mathematics - Class 11 JEEM21Ma18fc Solution: The number of the real roots of the equation is ________.

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

Step-by-Step Verified Solution

When x > 5

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

⇒x2+3x−4=274\Rightarrow {x^2} + 3x - 4 = {{27} \over 4}

⇒x2+3x−434=0\Rightarrow {x^2} + 3x - {{43} \over 4} = 0

⇒4x2+12x−43=0\Rightarrow 4{x^2} + 12x - 43 = 0

x=−12±144+6888x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}

x=−12±8328=−12±28.88x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}

=−3+7.22= {{ - 3 + 7.2} \over 2}

=−3+7.22,−3−7.22= {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}

= 2.1, -5.1 [ both are rejected as x should be > 5 ]

(Therefore no solution)

For x≤5x \le 5

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

x2+x+6−274=0{x^2} + x + 6 - {{27} \over 4} = 0

4x2+4x−3=04{x^2} + 4x - 3 = 0

x=−4±16+488x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}

x=−4±88⇒x=−128,48x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}

∴\therefore So, the equation have two real roots.

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