JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, equation (x+1)2+|(x−5)|=‌

27

4

Case I For x≥5
⇒‌‌(x+1)2+(x−5)‌‌=‌

27

4

⇒‌‌x2+3x−4‌‌=‌

27

4

⇒‌‌4x2+12x−43‌‌=0
x‌‌x‌‌=‌

−12±√144+688

8

∴‌‌‌‌=‌

−12±√832

8

‌‌=‌

−12±28.8

8

x‌‌=‌

−3±7.2

2

x‌‌=‌

−3+7.2

2

,‌

−3−7.2

2

Both the values are less than 5.
∴ No solution from here.
Case II x<5
⇒‌‌‌‌(x+1)2−(x−5)=‌

27

4

⇒‌‌x2+x+6−‌

27

4

=0
⇒‌‌4x2+4x−3=0
⇒‌‌x=‌

−4±√16+48

8

⇒‌‌x=‌

−12

8

,‌

4

8

,‌ both are less than ‌5.
∴ These values must be the solution.
Hence, here 2 real roots are possible.