In the real number system, the equation | vedclass

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Question

(d)

We have,

$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$

$\Rightarrow \sqrt{(\sqrt{x-1})^2-2(2) \sqrt{x-1}+(2)^2}$

$+\sqrt{(\sqrt{

Vedclass · Accepted Answer

infinitely many solutions

Vedclass · Answer

(d)

We have,

$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$

$\Rightarrow \sqrt{(\sqrt{x-1})^2-2(2) \sqrt{x-1}+(2)^2}$

$+\sqrt{(\sqrt{x-1})^2-2 	imes 3 \sqrt{x-1}+(3)^2}=1$

$\Rightarrow \quad \sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1$

$\Rightarrow \quad|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$

$x \in[5,10]$

$\sqrt{x-1}-2-\sqrt{x-1}+3=1$

$	herefore$

Hence, $x$ has infinite solutions in $x \in[5,10]$.

In the real number system, the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has

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