x = 1 is not an extraneous solution.
✓(x² + 3) = x + 1
Taking square on both the sides of the equation, we get
(✓(x² + 3))² = (x + 1)²
x² + 3 = (x + 1)²
Using the identity (a+b)²= a² + b² + 2ab on the right hand side of the equation, we get
x² + 3 = x² + (1)² + 2(x)(1)
x² + 3 = x² + 1 + 2x
2x = 2
x = 1
Confirming the value is extraneous solution or not
Putting the root x = 1 in the original equation ✓(x² + 3) = x + 1, we get
✓(1² + 3) = 1 + 1
✓4 = 2
2 = 2
⇒ LHS = RHS
∴ x = 1 is not an extraneous solution.
Know more about Extraneous Solutions: -brainly.com/question/20532561
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