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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

Arun
23328 Points
```							The roots are x1>0 and x2Write 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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