Quadratic Equation Question 9
Question 9 - 31 January - Shift 2
The equation
$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0, x \in R$ has:
(1) two solutions and both are negative
(2) no solution
(3) four solutions two of which are negative
(4) two solutions and only one of them is negative
Show Answer
Answer: (1)
Solution:
Formula: Roots of equations
$ e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0 $
Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$
Now, $\mathrm{t}^4+8 \mathrm{t}^3+13 \mathrm{t}^2-8 \mathrm{t}+1=0$
Dividing equation by $\mathrm{t}^2$,
$ \begin{aligned} & \mathrm{t}^2+8 \mathrm{t}+13-\frac{8}{\mathrm{t}}+\frac{1}{\mathrm{t}^2}=0 \\ & \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+8\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)+13=0 \\ & \left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)^2+2+8\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)+13=0 \end{aligned} $
Let $\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}$
$ \begin{aligned} & z^2+8 z+15=0 \\ & (z+3)(z+5)=0 \\ & z=-3 \text { or } z=-5 \end{aligned} $
So, $t-\frac{1}{t}=-3$ or $t-\frac{1}{t}=-5$ $\mathrm{t}^2+3 \mathrm{t}-1=0$ or $\mathrm{t}^2+5 \mathrm{t}-1=0$
$\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2}$ or $\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}$ as $\mathrm{t}=\mathrm{e}^{\mathrm{x}}
So $\mathrm{t}$ must be positive,$\mathrm{t}=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}$
So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$
$ \text { or } \mathbf{x}=\ln \left(\frac{\sqrt{29}-5}{2}\right) $
Hence two solution and both are negative.
