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Q.
The equation
$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1 \text { has }$
Complex Numbers and Quadratic Equations
Solution:
Put $\sqrt{x-1}=t $
$\Rightarrow x-1=t^{2}$
or $x=t^{2}+1$
The given equation reduces to
$\sqrt{t^{2}+1+3-4 t}+\sqrt{t^{2}+1+8-6 t}=1$
where $t \geq 0$.
$\Rightarrow |t-2|+|t-3|=1$
where $t \geq 0$.
This equation will be satisfied if $2 \leq t \leq 3$.
Therefore, $2 \leq \sqrt{x-1} \leq 3$
or $5 \leq x \leq 10$.
$\therefore $ The given equation is satisfied for all values of $x$ lying in $[5,10]$.