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How many real roots does the equation have?

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SherzodAzamov
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How many real roots does the equation have? [#permalink] New post   13 Jan 2021, 21:32
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00:00
A
B
C
D
E

Difficulty:

85% (hard)

Question Stats:

38% (01:40) correct 62% (01:45) wrong based on 93 sessions
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How many real roots does the equation, \(3^{-x }= 4 + x - x^2\), have?

A. 0
B. 1
C. 2
D. 3
E. 4
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Official Answer
C
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HinalAmin24
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Re: How many real roots does the equation have? [#permalink] New post   15 Jan 2021, 11:37
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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



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
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MaryKaaviya
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How many real roots does the equation have? [#permalink] New post   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
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AVARGUR
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Re: How many real roots does the equation have? [#permalink] New post   29 Jun 2023, 22:04
I solved the question by drawing graphs of both equations, I don't know if there's any other method to solve the question.
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Diablobuster
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Re: How many real roots does the equation have? [#permalink] New post   02 Jul 2023, 18:24
Take determinant of the RHS of the equation. b^2 - 4ac > 0, so there are two distinct real roots. Now, if you divide that quadratic by 3^(-x), for all values of x, 3^(-x) is positive. Therefore, if you subtract it from the LHS of the equation it has no real effect on the quadratic other than decreasing its overall value. Therefore, intuitively, the answer is C. Two real roots.
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Re: How many real roots does the equation have? [#permalink]
02 Jul 2023, 18:24
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