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

`rArrsqrt(3^2-4(3)+3)+sqrt(3^2-9)`

`rArrsqrt(9-12+3)+sqrt(9-9)`

`rArrsqrt0+sqrt0`

⇒ 0

∴ LHS = 0

Similarly, substitute x = 3 in RHS.

`rArrsqrt(4(3)^2-14(3)+16)`

`rArrsqrt(4xx9-42+16)`

`rArrsqrt(36-42+16)`

`rArrsqrt(52-42)`

`rArrsqrt10`

∴ RHS `=sqrt10`

Now, we can observe that

LHS ≠ RHS

∴ x = 3 is not a solution or root for the equation.