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How many real roots does the equation have?
[#permalink]
13 Jan 2021, 21:32
13 Jan 2021, 21:32
25
Bookmarks
Show timer
00:00
A
B
C
D
E
Difficulty:
95%
(hard)
Question Stats:
39% (01:51) correct
61%
(01:52)
wrong
based on 119
sessions
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
Re: How many real roots does the equation have?
[#permalink]
15 Jan 2021, 11:37
15 Jan 2021, 11:37
2
Bookmarks
SherzodAzamov wrote:
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
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
How many real roots does the equation have?
[#permalink]
19 Jan 2021, 10:25
19 Jan 2021, 10:25
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
