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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Updated on: 10-Oct-2022

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