Solution of the equation sqrt {x + 3 - 4 | vedclass

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Question

Let $u=x-1$. We can then rewrite the left hand side of the equation as

$\sqrt{u+4-4 \sqrt{u}}+\sqrt{u+9-6 \sqrt{u}}$

$=\sqrt{(\sqrt{u}-2)^{2}}+\

Vedclass · Accepted Answer

$x \in \left[ {5,10} \right]$

Vedclass · Answer

Let $u=x-1$. We can then rewrite the left hand side of the equation as

$\sqrt{u+4-4 \sqrt{u}}+\sqrt{u+9-6 \sqrt{u}}$

$=\sqrt{(\sqrt{u}-2)^{2}}+\sqrt{(\sqrt{u}-3)^{2}}$

$=|\sqrt{u}-2|+|\sqrt{u}-3|$

Note the presence of $\sqrt{u}$ in the equation and that we are only looking for real values, so we have the restriction $u \geq 0$. With that, we will now consider all remaining cases:

Case $1: 0 \leq u \leq 4$

$|\sqrt{u}-2|+|\sqrt{u}-3|=1$

$\Rightarrow 2-\sqrt{u}+3-\sqrt{2}=1$

$\Rightarrow-2 \sqrt{u}=-4$

$\Rightarrow \sqrt{u}=2$

$\Rightarrow u=4$

Thus $u=4$ is the only solution in the interval [0,4]

Case $2: 4 \leq u \leq 9$

$|\sqrt{u}-2|+|\sqrt{u}-3|=1$

$\Rightarrow \sqrt{u}-2+3-\sqrt{u}=1$

$\Rightarrow 1=1$

As this is a tautology, every value in [4,9] is a solution.

Case $3: u \geq 9$

$|\sqrt{u}-2|+|\sqrt{u}-3|=1$

$\Rightarrow \sqrt{u}-2+\sqrt{u}-3=1$

$\Rightarrow 2 \sqrt{u}=6$

$\Rightarrow \sqrt{u}=3$

$\Rightarrow u=9$

Thus $u=9$ is the only solution in the interval $[9, \infty)$

Taken together, we have [4,9] as the solution set for real values of $u$. Substituting in $x=u+1,$ we arrive at the final solution set $x \in[5,10]$

Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is

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