Exam Question some confusion?

Asked Oct 01 '22 at 16:33

Active Oct 01 '22 at 16:33

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Given $$ y=x+\sqrt{ x+ \sqrt {x + \sqrt {x +.....}} } $$

Then y(2)=?

A)4 only B)1 only C) 4 or 1 D) undefined..

My answer is

Taking root both side $$ {\sqrt y} = {\sqrt {x+\sqrt{ x+ \sqrt {x + \sqrt {x +.....}} }}} $$

$$ {\sqrt y} = {\sqrt {x + \sqrt y}} $$

$$ y=x+ \sqrt{y} $$

$$ x=2 $$ $$ y=2+\sqrt{y}$$ $$ y-2 = \sqrt{y} $$ $$ (y-2)^2 =y $$ $$ y^2 +4 -4y =y $$ $$ y^2 -5y+4=0$$

$$y=4 or 1 $$

But we know the value of the square root always greater than 0. $$y=2 +\sqrt{....}$$

so the value is always greater than 2.

my question is in the above solution where I did wrong so I get the value 1?

asked Oct 01 '22 at 16:33

Nitish

1

The value is 4. – Jean Marie Oct 01 '22 at 16:35
When you start squaring you may introduce spurious roots. Consider this example: Let $a=2$. Therefore $a^2=4$. Hence $a=\pm 2$. – ancient mathematician Oct 01 '22 at 16:38
@ancient mathematician : It is true, but I don't think this is the problem here. – Jean Marie Oct 01 '22 at 16:49
@JeanMarie Are you sure. I think the line $y-2=\sqrt{y}$ is correct, and only has the root $y=4$, but on squaring we lose precision and pick up the extra (spurious) root $1$. – ancient mathematician Oct 02 '22 at 06:33
@ancient mathematician My point is that all spurious roots can be eliminated afterwards. – Jean Marie Oct 02 '22 at 13:04

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