Question

Determine, if 3 is a root of the equation given below :
$\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+16}$

Answer 1

Given to check whether 3 is a root of the equation

$\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+16}$

Here,

$LHS=\sqrt{x^2-4x+3}+\sqrt{x^2-9}$ and $RHS=\sqrt{4x^2-14x+16}$

Substitute $x=3$ in LHS

$\Rightarrow \sqrt{3^2-4\left(3\right)+3}+\sqrt{3^2-9}$

$\Rightarrow \sqrt{9-12+3}+\sqrt{9-9}$

$\Rightarrow \sqrt{0}+\sqrt{0}=0$

$\therefore LHS=0$

Similarly substitute $x=3$ in RHS

$\Rightarrow \sqrt{4(3{)}^2-14(3)+16}$

$\Rightarrow \sqrt{4\times 9-42+16}$

$\Rightarrow \sqrt{36-42+16}$

$\Rightarrow \sqrt{52-42}=\sqrt{10}$

$\therefore RHS=\sqrt{10}$

Now, we can observe that

$LHS\ne RHS$

$\therefore x=3$ is not a solution or root for the equation

$\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+16}$