The equation e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e | vedclass

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

Question

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0$

Let $e^x=t$

Now, $t^4+8 t^5+13 t^2-8 t+1=0$

Dividing equation by $t ^2$,

$t^2+8 t+13-\frac{8}{t}+

Vedclass · Accepted Answer

two solutions and both are negative

Vedclass · Answer

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0$

Let $e^x=t$

Now, $t^4+8 t^5+13 t^2-8 t+1=0$

Dividing equation by $t ^2$,

$t^2+8 t+13-\frac{8}{t}+\frac{1}{t^2}=0$

$t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}ight)+13=0$

$\left(t-\frac{1}{t}ight)^2+2+8\left(t-\frac{1}{t}ight)+13=0$

Let $t-\frac{1}{t}=z$

$z^2+8 z+15=0$

$(z+3)(z+5)=0$

$z=-3 	ext { or } z=-5$

So, $t -\frac{1}{ t }=-3$ or $t -\frac{1}{ t }=-5$

$t^2+3 t-1=0 	ext { or } t^2+5 t-1=0$

$t=\frac{-3 \pm \sqrt{13}}{2} 	ext { or } t =\frac{-5 \pm \sqrt{29}}{2}$

as $t = e ^{ x }$ so $t$ must be positive,

$t=\frac{\sqrt{13}-3}{2} 	ext { or } \frac{\sqrt{29}-5}{2}$

So, $x=\ln \left(\frac{\sqrt{13}-3}{2}ight)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}ight)$

Hence two solution and both are negative.

The equation $e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in R$ has:

Similar Questions

If for a posiive integer $n$ , the quadratic equation, $x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .\;.\;.\; + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$ has two consecutive integral solutions, then $n$ is equal to:

Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to ............

If $\alpha, \beta $ and $\gamma$ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$, then ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ is equal to

The least integral value $\alpha $ of $x$ such that $\frac{{x - 5}}{{{x^2} + 5x - 14}} > 0$ , satisfies

The number of real solutions of the equation $x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$ is