The equation e^{4x}+8e³x}+13e²x}-8e^{x}+1=0, x {R} has: [31-Jan-2023 Shift 2]

Correct Answer is : two solutions and both are negative

two solutions and both are negative

four solutions two of which are negative

two solutions and only one of them is negative

Correct Answer is : two solutions and both are negative

Test Index

Quadratic Equation and Inequalities Part 2

Section: Mathematics

Question : 64 of 98

Marks: +1, -0

The equation

e^{4x}+8e^{3x}+13e^{2x}-8e^{x}+1=0, x \in \mathbb{R}

Solution:

e^{4x}+8e^{3x}+13e^{2x}-8e^{x}+1=0

Let

e^{x}=t

Now,

t^4+8t^3+13t^2-8t+1=0

Dividing equation by

t^2

\;t^2+8t+13-\;\frac{8}{t}+\;\frac{1}{t^2}=0

\;t^2+\;\frac{1}{t^2}+8\left(t-\;\frac{1}{t}\right)+13=0

\;\left(t-\;\frac{1}{t}\right)^2+2+8\left(t-\;\frac{1}{t}\right)+13=0

Let

t-\;\frac{1}{t}=z

\;z^2+8z+15=0

\;(z+3)(z+5)=0

\;z=-3 \;\text{ or }\; z=-5

So,

t-\;\frac{1}{t}=-3

t-\;\frac{1}{t}=-5

\;t^2+3t-1=0 \;\text{ or }\; t^2+5t-1=0

\;t=\;\frac{-3 \pm \sqrt{13}}{2} \;\text{ or }\; t=\;\frac{-5 \pm \sqrt{29}}{2}

t=e^{x}

t

must be positive,

t=\;\frac{\sqrt{13}-3}{2} \;\text{ or }\; \frac{\sqrt{29}-5}{2}

So,

x=\ln\left(\frac{\sqrt{13}-3}{2}\right)

x=\ln\left(\frac{\sqrt{29}-5}{2}\right)

Hence two solution and both are negative.

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