Algebra

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16 Jul 2009 21:12:38 IST

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plz solve the following & give me solution

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then the value of x = ?

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Joined: 10 Jul 2008

Posts: 598

17 Jul 2009 09:06:02 IST

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The given equation
$\sqrt{a+\sqrt{a+x}}=x$ .
Let $\sqrt{a+x}=y$ . Then we get a set of equations
$\sqrt{a+y}=x$ and $\sqrt{a+x}=y$
Squaring them we obtain
$a+y=x^2$ and $a+x=y^2$
Subtracting the second from the first give us
$y-x=x^2-y^2=(x+y)(x-y)\quad\Rightarrow\ (x-y)(x+y+1)=0$
Now there are two possibilities:
i) If $x-y=0$ then $y=x$ , hence we get
$x^2-x-a=0$
which means the roots are
$x_{1,2}=\dfrac{1\pm \sqrt{1+4a}}{2}$

(ii) If $x+y+1=0$ , then we have $y=-1-x$ and hence we get
$x^2+x+1-a=0$
which gives
$x_{3,4}=\dfrac{-1\pm\sqrt{4a-3}}{2}$
These possibilities for the roots must be checked for extraneous roots.

Forum Expert

Joined: 10 Jul 2008

Posts: 598

17 Jul 2009 09:29:52 IST

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If we require only the positive roots (as $\sqrt{z}\geq 0$ for a real $z$ ), then it is easy to ascertain that the only such value of $x$ will be
$x_1=\dfrac{1+\sqrt{1+4a}}{2}$

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