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If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi

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Bunuel
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If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 01:53
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A
B
C
D
E

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56% (01:56) correct 44% (02:17) wrong based on 144 sessions
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If \((3 + 2√2)^{(x^2 - 3)} + (3 - 2√2)^{(x^2 - 3)} = b\) which of the following can be the value of b?

A. -1
B. 0
C. 1
D. \(\sqrt{2}\)
E. 2


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E
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 03:00
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If \(x=\sqrt3\),

\(b=(3+2\sqrt2)^0 + (3-2\sqrt2)^0=1+1=2\)

Answer is (E)

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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 03:36
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IMO E

Let x=√ 3 (Since no information about x is available so we can assume it to be anything)
x^2-3 = 0

(3+2√2)^(x2−3)+(3−2√2)^(x2−3)=b
=> (3+2√2)^0 + (3+2√2)^0 = b
=> 1 + 1 = b
=> b =2

E. 2
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 03:42
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\(b=(3+2\sqrt2)^{x^2-3} + (3-2\sqrt2)^{x^2-3}\)

Also,
\((3+2\sqrt2)*(3-2\sqrt2) = 9-8=1\)

\((3+2\sqrt2) = \frac{1}{(3-2\sqrt2)}\)

or

\((3+2\sqrt2)^{x^2-3} = \frac{1}{(3-2\sqrt2)^{x^2-3}}\)

Sum of a positive number and it's reciprocal can never be less than 2.

E
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 03:44
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If \((3+2√2)^{(x^2−3)} + (3−2√2)^{(x^2−3)} = b\) which of the following can be the value of b?

A. -1
B. 0
C. 1
D. √2
E. 2

Since 3+2√2 > 0 & 3-2√2 > 0, b ≤ 0. So, b > 0
A and B are out.

As question asks b can take a value out of the five options, we need to find one options that satisfies the condition.
Hence the easiest solution is when \((xˆ2-3) = 0\)(there's no condition for x which can take any value), Thus, \((3+2√2)^{(x^2−3)}\) = 1 and
\((3-2√2)^{(x^2−3)}\) = 1 also

So, \((3+2√2)^{(x^2−3)} + (3+2√2)^{(x^2−3)} = 1+1 = 2\)

Answer E.
Note: Many answers are possible i.e. b can take any value depending the value we consider for 'x' but since 2 is available as answer and in PS only one option is correct, we need to bother whether C and D are correct.
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 04:18
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Is it E ?

If (3+2√2)^(x^2−3)+(3−2√2)^(x^2−3)=b
which of the following can be the value of b?


if x^2−3 = 0 then total expression can be
(3+2√2)^(0)+(3−2√2)^(0)= 1 + 1 = 2 .
Hence E .
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 04:44
Either you can go ahead and start solving forgo with a smart approach.
Put x = sqrt(3)
so x^2-3 = 0

the equation is reduced to (....)^0 + (.....)^0 = 1+1 = 2
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 06:08
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Quote:
If (3+2√2)(x2−3)+(3−2√2)(x2−3)=b which of the following can be the value of b?

A. -1
B. 0
C. 1
D. 2√2
E. 2


if x=√3, then (..)^0+(..)^0=1+1=2=b

ans (E)
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 08:43
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We have to concentrate on value of x2-3, this equals to 0, when x=+_/3, with this both x + Y = 1+1= 2= b

So answer is E

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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 13:11
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IMO E.

Lets consider the equation of the form (p+q)^r + (p-q)^r = b

As the Lets try with r= 0, 1

r=0 (i.e. x= sqrt(3)): we get 1 + 1 = b = 2 (which is in the options)
So that is the answer.

If that would not have been among the options, putting r=1; b = 6.

Looking forward to other solutions :)
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Re: If (3 + 2√2)^(x^2 - 3)} + (3 - 2√2)^(x^2 - 3) = 1 which of the followi [#permalink] New post   01 Jun 2020, 22:17
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(a+c)^d+(a-c)^d
we can have d=2,1,0...
x^2-3=1,
x=+2,-2
but , when d=1, b=6
when d=0, b=2

In the question they are asking can be the value of b, so 2 can be the value

Ans E
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Re: If (3 + 22)^(x^2 - 3)} + (3 - 22)^(x^2 - 3) = 1 which of the followi [#permalink] New post   10 Mar 2024, 07:21
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