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the equation 2^2x+(a-1)2^(x+1) + a = 0 has roots of opposite signs, then the set of values of a is ?
one year ago

Answers : (1)

Arun
23328 Points
							
The roots are x1>0 and x2
Write y=2^x, then y1 > 1 and 0

The quadratic equation 
y^2+(a-1)y+(1+a)=0 
has solutions 
y1,y2 = (1/2)(-(a-1) +/- D) = (1/2)(1-a +/- D) 
where 
D^2 = (a-1)^2-4*1*(a+1) 
=a^2-2a+1-4a-4 
=a^2-6a-3. 

Condition (1) : y1>1 
We require 
y1 = (1/2)(1-a + D) > 1 
D > 2+a-1 = 1+a 
D^2 = a^2-6a-3 > (1+a)^2 = 1+2a+a^2 
-8a > 4 
a

Condition (2) : y2 > 0. 
We require 
y2 = (1/2)(1-a - D) > 0 
1-a > D 
(1-a)^2 = a^2-2a+1 > D^2 = a^2-6a-3 
4a > -4 
a > -1. 

Condition (3) : y2
We require 
y2 = (1/2)(1-a - D)
1-a-D
-(1+a)
[-(1+a)]^2 = 1+2a+a^2
8a
a

Combining the 3 conditions, we have the ANSWER that the roots have opposite signs for values of a in the range -1
one year ago
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