Question

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

• A. four solutions two of which are negative
• B. two solutions and both are negative
• C. no solution
• D. two solutions and only one of them is negative

Answer 1

Answer: (B)

two solutions and both are negative

Answer 2

e4x+8e3x+13e2x−8ex+1=0
Let ex=t
Now, t4+8t3+13t2−8t+1=0
Dividing equation by t2,
t2+8t+13−t8​+t21​=0
t2+t21​+8(t−t1​)+13=0
(t−t1​)2+2+8(t−t1​)+13=0
Let t−t1​=z
z2+8z+15=0
(z+3)(z+5)=0
z=−3 or z=−5
So, t−t1​=−3 or t−t1​=−5
t2+3t−1=0 or t2+5t−1=0
t=2−3±13​​ or t=2−5±29​​
as t=ex so t must be positive,
t=213​−3​ or 229​−5​
So, x=ln(213​−3​) or x=ln(229​−5​)
Hence two solution and both are negative.
So , the correct option is (B) : two solutions and both are negative