The number of real solutions of the equation | vedclass

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Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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.

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

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