Question

In the following, determine whether the given values are solutions of the given equation or not:

`sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16), x = 3`

Answer 1

Given: `sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16), x = 3`

x² – 4x + 3 = 9 – 12 + 3 = 0 

⇒ `sqrt(0) = 0`

x² – 9 = 9 – 9 = 0

⇒ `sqrt(0) = 0`

LHS = 0 + 0 = 0

4x² – 14x + 16 = 36 – 42 + 16 = 10

⇒ RHS = `sqrt(10) (≈ 3.1623)`

`0 ≠ sqrt(10)`; also the radicands are defined at x = 3 (x² – 9 = 0), so substitution is valid.

x = 3 is not a solution of the equation.