JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Basic Mathematics

The number of the real roots of the equation

(x + 1)^{2} + ∣ x - 5∣ = \frac{27}{4}

Answer: 2

Solution

For

x \geq 5

the equation reads

(x + 1)^{2} + x - 5 = \frac{27}{4}

, and its left side already equals

36

x = 5

and increases beyond, so there is no root on that side. For

x < 5

it reads

(x + 1)^{2} + 5 - x = \frac{27}{4}

, that is

x^{2} + x + 6 = \frac{27}{4}

, or

4 x^{2} + 4 x - 3 = 0

, which factors as

(2 x + 3) (2 x - 1) = 0

and gives

x = - \frac{3}{2}

and

x = \frac{1}{2}

, both admissible. The equation therefore has exactly

2

real roots.

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer