Question

In exercises 18 and 19 determine which solution if any is an extraneous solution 18.sprt(3x-2)=x; x=1,x=2. 19. Sprt(x+6=x; x=3,x=-2

Answer 1

18. No extraneous solution.

19. The extraneous solution is x=-2

Step-by-step explanation

18. The given radical equation is:

`√(3x - 2) = x`

Solving this radical equation yields

`x=1,x=2`

We check for an extraneous solution by substituting each value into the equation.

Checking for x=1,

`√(3 * 1 - 2) = 1`

`√(3- 2) = 1`

`√(1) = 1`

`1 = 1`

This is true.

Checking for x=2

`√(3 * 2- 2) = 2`

`√(6- 2) = 2`

`√(4) = 2`

`2 = 2`

This is also true. Hence there is no extraneous solution.

19. The given radical equation is:

`√(x + 6) = x`

Solving this equation yields,

`x=3,x=-2`

Checking for x=3,.

`√(3+ 6) = 3`

`√(9) = 3`

3=3.

This is a true solution.

Checking for x=-2.

`√( - 2 + 6) = - 2`

`√(4) = - 2`

`2 \\e - 2`

Hence x=-2 is an extraneous solution.