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13 Jan 2021, 21:32
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (01:56) correct
62%
(01:48)
wrong
based on 165
sessions
History
Date
Time
Result
Not Attempted Yet
How many real roots does the equation, \(3^{-x }= 4 + x - x^2\), have?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
15 Jan 2021, 11:37
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SherzodAzamov
3^(-x) = 4 + x - x^2
- 3^(-x) = x^2 - x - 4
- 3^(-x) = (x - 3)(x + 1)
So x - 3 = - 3^(-x) or x + 1 = - 3^(-x)
2 roots
19 Jan 2021, 10:25
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Answer is C
3^-x=4+x-x^2 ==>x^2-x-4+3^-x=0
[align=]
By formula roots=\(-b+/-\frac{\sqrt{b^2-4ac} }{2a}\)[/align]
thus roots are
\(1+\frac{\sqrt{17}}{2} \),\(1-\frac{\sqrt{17}}{2}\)
Ans = C
3^-x=4+x-x^2 ==>x^2-x-4+3^-x=0
[align=]
By formula roots=\(-b+/-\frac{\sqrt{b^2-4ac} }{2a}\)[/align]
thus roots are
\(1+\frac{\sqrt{17}}{2} \),\(1-\frac{\sqrt{17}}{2}\)
Ans = C
