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

\(\sqrt{\frac{3}{2}+x} = x\); 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

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

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)

Posted from my mobile device

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.

√(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

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 [-b +/- sqrt(b^2 - 4ac)] / 2a

Therefore [-(-1) + /- sqrt[(-1)^2 - (4*1*-3/2)]] / 2

= [1 +sqrt(7)] / 2 and [1 - sqrt(7)]/2

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


Therefore [1 +sqrt(7)] / 2

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 \(\sqrt{3/2}=\sqrt{1.5}>1\)
\(\sqrt{1.5+{\sqrt{1.5}}<x........x>\sqrt{1.5+1}=\sqrt{2.5}\), 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

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

The answer is E.

Posted from my mobile device

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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.

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

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