Question

The number of real solutions of the equation |$x^{2} + 2x + 2 = 0$ + 4x + 3| + 2x + 5 = 0 are.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Here two cases arise viz.

Case I : $\alpha$

This gives $\beta$

⇒$\alpha$ ⇒ $\gamma$ ⇒ $\alpha$

$\delta$ is not satisfying the condition $a,b$, so $c$ is the only solution of the given equation.

Case II : $\alpha + \alpha^{2}$

This gives –$a + 2b + c$

⇒$\left| 3x^{2} + 12x + 6 \right| = 5x + 16$

⇒$x + 2 > \sqrt{x + 4}$

⇒ $x < - 2$

Hence $x > 0$ satisfy the given condition

$- 3 < x < 0$, while $- 3 < x < 4$is not satisfying the condition. Thus number of real solutions are two.