If x is a positive number and sqrt(3/2 + (sqrt(3/2 + sqrt(3/2 +

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Competition Mode Question

If  x  is a positive number and  \sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x , where the given expression extends to an infinite number of roots, then what is the value of  x ?

A.  \frac{1-\sqrt{7}}{2}

B.  \frac{1}{2}

C.  \frac{\sqrt{7}-1}{2}

D.  \frac{\sqrt{7}}{2}

E.  \frac{1+\sqrt{7}}{2}

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Bunuel · Accepted Answer

GMATBusters · Answer

This is a very good type of question: please see the approach to solve the question https://gmatclub.com/forum/download/file.php?id=53226 WhatsApp Image 2020-03-06 at 2.53.20 PM.jpeg

yashikaaggarwal · Answer

Whole √3/2+ √3/2 +...... = X
Will become √3/2+X = X
Put options one by one. 
Only 1-√7/2 is satisfying the equation. So OA is A

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rohitguglani · Answer

It can be written as :

\sqrt{(3/2 + x)} = x 
Solving it becomes

-->  3/2 + x = x^2

-->  2x^2-2x-3 =0

--> x =  (1+\sqrt{7})/2  or  (1-\sqrt{7}/2)  ( This cant be possible since x is positive) . Hence  (1+\sqrt{7}/2)  is the answer

E

freedom128 · Answer

x = √(3/2 +x) must be >=0
x^2 = 3/2 +x
x^2 -x -3/2 = 0

Solution: 
x1= (1-√7)/2 (N.A, it's negative) 
x2= (1+√7)/2 (Yes!)

FINAL ANSWER IS (E)

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unraveled · Answer

If x is a positive number and  \sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x , where the given expression extends to an infinite number of roots, then what is the value of x?

A.  \frac{1−√7}{2}

B.  \frac{1}{2}

C.  \frac{√7−1}{2}

D.  \frac{√7}{2}

E.  \frac{1+√7}{2}

As x > 0 
Looking the options we have A is out since it is negative.

\sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x

So it can be written as 
 \sqrt{\frac{3}{2}+x}=x 
Squaring both sides 
 \frac{3}{2} + x = x^2 
 2x^2 - 2x - 3 = 0 
x =  \frac{1+√7}{2}  and  \frac{1−√7}{2}

So x =  \frac{1+√7}{2}

Answer E.

ArunSharma12 · Answer

\sqrt{\frac{3}{2}+x} = x ; as x extends to infinite terms

\frac{3}{2} + x = x^2

x^2-x-\frac{3}{2}=0

use the formula,  = \frac{-b ±\sqrt{b^2- 4*a*c}}{2a}

x = \frac{1 ± \sqrt{1+6}}{2}

x = \frac{1 + \sqrt{7}}{2}  (x can not be negative)

Ans: E

eakabuah · Answer

x=√(3/2+√(3/2+√(3/2+...
Hence x=√(3/2+x)
From this we know that x>√(3/2)
x^2=3/2+x
x^2-x-3/2=0
x=(1±√(1+4*1*3/2))/2
x=(1±√7)/2
so x=(1+√7)/2 or x=(1-√7)/2
But (1-√7)/2 is negative and x cannot be negative.
Hence x=(1+√7)/2

The answer is E.

Jawad001 · Answer

√(3/2 + x) = x
 Or, (3/2 + x) = x^2
Or , 2x^2 - 2x -3 = 0

X = (-(-2)±√(〖(-2)〗^2-4*2*-3))/(2*2)
X = (2±√(2^2+24))/4
X = (2±√(4^ +24))/4
X = (1±√7)/2
Since x is positive, x = (1+√7)/2
Answer: E

arunkumar1975 · Answer

Option E

we can rewrite the equation as x = sqrt(3/2 + x)

squaring both sides, we get x^2 = 3/2 + x

Therefore x^2 - x - 3/2 = 0

The roots of the quadratic equation is   / 2a

Therefore  ] / 2

=   / 2 and  /2

We take the positive value as the original equation of x represents a positive value.

Therefore   / 2

chetan2u · Answer

If x is a positive number and \sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x , where the given expression extends to an infinite number of roots, then what is the value of x ? Two ways.. If you are aware of the type of these questions.. \sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x .. We can further write this as \sqrt{\frac{3}{2}+x}=x Square both sides... \frac{3}{2}+{x}=x^2......x^2-x-\frac{3}{2}=0 Apply the formula for roots, = \frac{-b ±\sqrt{b^2- 4*a*c}}{2a}=\frac{1^2 ± \sqrt{(-1)^2+4*\frac{3}{2}}}{2} = \frac{1 ± \sqrt{1+6}}{2} But x is surely positive, so x= \frac{1 + \sqrt{7}}{2} Next, if you do not know anything but are looking for a start... Now 3/2=1.5 what is \sqrt{3/2}=\sqrt{1.5}>1 \sqrt{1.5+{\sqrt{1.5}}\sqrt{1.5+1}=\sqrt{2.5} , so x is surely >1.5 Look at the choices that gives you these values.. A. \frac{1-\sqrt{7}}{2} ... a NEGATIVE value...N) B. \frac{1}{2} =0.5...NO C. \frac{\sqrt{7}-1}{2} .... \sqrt{7} is between 2 and 3. Even if it is 3, \frac{3-1}{2}=1 ..NO D. \frac{\sqrt{7}}{2} ....Surely <\frac{3}{2}...<1.5 ....NO E. \frac{1+\sqrt{7}}{2} ....Surely <\frac{1+3}{2}...<2 ......Possible So, Only E is possible

lacktutor · Answer

If you square the both sides of the given expression, we can get 
—>  x^{2} —x —\frac{3}{2 = 0  
( quadratic equation) 
 x_1 =  1–\sqrt{7} /2} 
 x_2 = \frac{1+ \sqrt{7}}{2} 
As x is positive, the value of x should be  \frac{1+\sqrt{7} }{2}

The answer is E.

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NarayanaGupta007 · Answer

after solving Eq Option A & E both are roots but x can be only positive under root so A rejected and E is the answere
Option E has + value of X

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=W3V2vt_vqMw

Given that  \sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x

Now the expression inside the square roots extends till infinity and if we note then the terms are repeating.
So, we can approximate the problem by writing terms after the first  \frac{3}{2}  = x

=>  \sqrt{\frac{3}{2}+ x}=x

Squaring both the sides
 \frac{3}{2}  + x =  x^2 
=>  2x^2 - 2x - 3 = 0 
=> x =  \frac{-(-2) ± \sqrt{(-2)^2 - 4 * 2 *-3}}{2*2}  =  \frac{2 ± \sqrt{28}}{4} 
=  \frac{2 ± 2\sqrt{7}}{4}  =  \frac{1 ± \sqrt{7}}{2}

So,  Answer will be E 
Hope it helps!

Watch the following video to MASTER Roots

https://www.youtube.com/watch?v=5oAec-Y6kOQ

If x is a positive number and sqrt(3/2 + (sqrt(3/2 + sqrt(3/2 +

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GMAT Problem Solving (PS) Questions

If x is a positive number and sqrt(3/2 + (sqrt(3/2 + sqrt(3/2 +

Prep Toolkit

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