x = 5 is not an extraneous solution of the equation.
✓(2x + 6) - ✓(x - 1) = 2
✓(2x + 6) = 2 + ✓(x - 1)
Squaring both the sides of equation, we get
(✓(2x + 6))² = (2 + ✓(x - 1))²
2x + 6 = (2)² + (✓(x - 1))² + 2(2)(✓(x - 1))
2x + 6 = 4 + x - 1 + 4(✓(x - 1))
x + 3 = 4✓(x - 1)
Squaring again on both the side of the equation, we get
(x + 3)² = (4✓(x - 1))²
x² + 3² + 2×3×x = 16(x - 1)
x² + 9 + 6x = 16x - 16
x² - 10x + 25 = 0
By applying middle term splitting on equation (1), we get
x² - 5x - 5x + 25 = 0
x(x - 5) -5(x - 5) = 0
(x - 5) (x - 5) = 0
x = 5 and 5.
Check for extraneous solutions by putting value of x in original equation -
✓(2×5 + 6) - ✓(5 - 1) = 2
✓16 - ✓4 = 2
4 - 2 = 2
2 = 2
LHS = RHS
x = 5 is not an extraneous solution of the equation.
Know more about roots of equation -
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