Determine, if 3 is a root of the equation given below:$\sqrt{x^2-4x+3} + \sqrt{x^2-9} = \sqrt{4x^2-14x+16}$
Given:
Given equation is $\sqrt{x^2-4x+3} + \sqrt{x^2-9} = \sqrt{4x^2-14x+16}$.
To do:
Here, we have to determine if 3 is a root of the given equation.
Solution:
For $x=3$
LHS
$\sqrt{x^2-4x+3} + \sqrt{x^2-9} = \sqrt{(3)^2-4(3)+3} + \sqrt{(3)^2-9}$
$=\sqrt{9-12+3}+\sqrt{9-9}$
$=0$
RHS
$\sqrt{4x^2-14x+16}=\sqrt{4(3)^2-14(3)+16}$
$=\sqrt{36-42+16}$
$=\sqrt{10}$
LHS$≠$RHS
Therefore, $x=3$ is not a solution of the given equation.
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