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If the expression [m]x=[square_root]6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}[
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01 Sep 2022, 10:16
01 Sep 2022, 10:16
Expert Reply
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Question Stats:
60% (00:54) correct
40% (02:05) wrong based on 15 sessions
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If the expression \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\) extends to an an infinite number of roots and converges to a positive number k, then approximately what is the value of k?
A. 1
B. 3
C. 5 + √5
D. 5 + √6
E. 5√6
A. 1
B. 3
C. 5 + √5
D. 5 + √6
E. 5√6
If the expression [m]x=[square_root]6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}[
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02 Sep 2022, 07:59
02 Sep 2022, 07:59
i think the question should be regarding the value of x not k.
if it is then after solving x gives 2 values - > 3,-2
so given from options is B)3.
if it is then after solving x gives 2 values - > 3,-2
so given from options is B)3.
Re: If the expression [m]x=[square_root]6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}[
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02 Sep 2022, 08:10
02 Sep 2022, 08:10
Expert Reply
It says that converges to k not that is the value.
\(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Squaring both sides is the best approach for such questions
\(x^2 = 6 + \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Now since we know this goes on infinitely, we can say that the part after the 6 in the right hand side of the equation is equal to the original x
Rewrite the equation as
\(x^2 = 6 + x\)
\(x^2 - x - 6 = 0\)
\(x^2 + 2x - 3x - 6 = 0\)
\((x+2)(x-3) = 0\)
x = -2 or 3
Since x is a square root expression, it cannot be negative so
\(x = 3\) is the only solution
\(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Squaring both sides is the best approach for such questions
\(x^2 = 6 + \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Now since we know this goes on infinitely, we can say that the part after the 6 in the right hand side of the equation is equal to the original x
Rewrite the equation as
\(x^2 = 6 + x\)
\(x^2 - x - 6 = 0\)
\(x^2 + 2x - 3x - 6 = 0\)
\((x+2)(x-3) = 0\)
x = -2 or 3
Since x is a square root expression, it cannot be negative so
\(x = 3\) is the only solution
