TASK 2
Part 1
2√(x+1)+7=5 is a radical equation with an extraneous solution.
Part 2
2√(x+1)+7=5, 2√(x+1)=-2, √(x+1)=-1, square both sides: x+1=1, x=0.
But 2√(x+1)+7=9 when x=0 and 9≠5. So x=0 is an extraneous solution.
2√(x+1)+5=7 is a radical equation with solution x=0.
Part 3
The extraneous solution is a result of squaring because (-1)²=1²=1. This creates an ambiguity. But √(x+1) specifically implies the positive root. This is the reason why the second equation works, because the positive root satisfies the equation.