The number of real solutions of the equation | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided

If the sum of all the roots of the equation $e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0$ is $\log _{ e } P$, then $p$ is equal to

Let $S$ be the set of all real roots of the equation, $3^{x}\left(3^{x}-1\right)+2=\left|3^{x}-1\right|+\left|3^{x}-2\right| .$ Then $\mathrm{S}$

The number of solution$(s)$ of the equation $2^x = x^2$ is

Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $P_n=\alpha^n-\beta^n, n \in N$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to :