# Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 625: 52

a) $y= 9$, extraneous--$y=-2$ b) $y= -2$, extraneous--$y=9$ c) The extraneous solution would be an acceptable (normal solution), while the normal solution would be the extraneous solution. Parts a) and b) show that the normal and extraneous answers swap places when -1 is multiplied by one side of an equation.

#### Work Step by Step

a) $\sqrt {7y+18} =y$ $(\sqrt {7y+18})^2 =y^2$ $7y+18 = y^2$ $y^2-7y-18 = 7y+18-7y-18$ $(y-9)(y+2) = 0$ $y-9 = 0$ $y=9$ $y+2=0$ $y=-2$ The square root of a number can't be negative, so $y=-2$ is extraneous. $\sqrt {7y+18} =y$ $\sqrt {7*9+18} =9$ $\sqrt {63+18} = 9$ $\sqrt {81} = 9$ $9 = 9$ b) $\sqrt {7y+18} =-y$ $(\sqrt {7y+18})^2 =(-y)^2$ $7y+18 = y^2$ $y^2-7y-18 = 7y+18-7y-18$ $(y-9)(y+2) = 0$ $y-9 = 0$ $y=9$ $y+2=0$ $y=-2$ The square root of a number can't be positive if the variable is multiplied by -1, so $y=9$ is extraneous.

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