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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)`
Solution
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(9-18+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
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