Number of real solutions of equation |{x^2} + 4x + 3| + 2x + 5 = 0 are

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4-2.Quadratic Equations and Inequations

hard

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

$1$

$2$

$3$

$4$

(IIT-1988)

Solution

(b) Here two cases arise viz.

Case I : ${x^2} + 4x + 3 > 0$

This gives ${x^2} + 4x + 3 + 2x + 5 = 0$

==>${x^2} + 6x + 8 = 0$==> $(x + 2)(x + 4) = 0$ ==> $x = – 2, – 4$

$x = – 2$ is not satisfying the condition ${x^2} + 4x + 3 > 0$, so $x = – 4$ is the only solution of the given equation.

Case II : ${x^2} + 4x + 3 < 0$

This gives -$({x^2} + 4x + 3) + 2x + 5 = 0$

==> $ – {x^2} – 2x + 2 = 0 \Rightarrow {x^2} + 2x – 2 = 0$

==>$(x + 1 + \sqrt 3 )(x + 1 – \sqrt 3 ) = 0$

==> $x = – 1 + \sqrt 3 , – 1 – \sqrt 3 $

Hence $x = – (1 + \sqrt 3 )$ satisfy the given condition ${x^2} + 4x + 3 < 0$, while $x = – 1 + \sqrt 3 $ is not satisfying the condition. Thus number of real solutions are two.

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