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11 Aug 2021, 06:15
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B
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Difficulty:
35%
(medium)
Question Stats:
64% (01:02) correct
36%
(01:15)
wrong
based on 205
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What is the value of \(\sqrt{6+\sqrt{6+ \sqrt{6+...}}}\) where the given expression extends to an infinite number of roots ?
A. 2
B. 3
C. 4
D. 6
E. 9
A. 2
B. 3
C. 4
D. 6
E. 9
11 Aug 2021, 06:26
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let the given expression be y
then, sq root(6+y)=y
6+y=y^2
y^-y-6=0
(y-3)(y+2)=0
y=3,-2
bt negative root is not possible.
so, y=3
B it is
then, sq root(6+y)=y
6+y=y^2
y^-y-6=0
(y-3)(y+2)=0
y=3,-2
bt negative root is not possible.
so, y=3
B it is
11 Aug 2021, 06:28
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Solution:
Let us assume \(x=\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
So, we need the value of \(x\).
We have \(x=\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
On squaring both sides, we have
\(⇒ x^2=6+\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
\(⇒ x^2=6+x \) (since we know \(\sqrt{6+\sqrt{6+\sqrt{6...}}}=x\))
\(⇒ x^2-x-6=0 \)
\(⇒x^2-3x+2x-6=0\)
\(⇒ x(x-3)+2(x-3)=0\)
\(⇒ (x-3)(x+2)=0\)
\(⇒x=3\) or \(x=-2\)
Hence the right answer is Option B.
Let us assume \(x=\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
So, we need the value of \(x\).
We have \(x=\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
On squaring both sides, we have
\(⇒ x^2=6+\sqrt{6+\sqrt{6+\sqrt{6...}}}\)
\(⇒ x^2=6+x \) (since we know \(\sqrt{6+\sqrt{6+\sqrt{6...}}}=x\))
\(⇒ x^2-x-6=0 \)
\(⇒x^2-3x+2x-6=0\)
\(⇒ x(x-3)+2(x-3)=0\)
\(⇒ (x-3)(x+2)=0\)
\(⇒x=3\) or \(x=-2\)
Hence the right answer is Option B.
