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

ANSWER

18. No extraneous solution.

19. The extraneous solution is x=-2

EXPLANATION

18. The given radical equation is:

$\sqrt{3x - 2} = x$

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,

$\sqrt{3 \times 1 - 2} = 1$

$\sqrt{3- 2} = 1$

$\sqrt{1} = 1$

$1 = 1$

This is true.

Checking for x=2

$\sqrt{3 \times 2- 2} = 2$

$\sqrt{6- 2} = 2$

$\sqrt{4} = 2$

$2 = 2$

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

19. The given radical equation is:

$\sqrt{x + 6} = x$

Solving this equation yields,

$x=3,x=-2$

Checking for x=3,.

$\sqrt{3+ 6} = 3$

$\sqrt{9} = 3$

3=3.

This is a true solution.

Checking for x=-2.

$\sqrt{ - 2 + 6} = - 2$

$\sqrt{4} = - 2$

$2 \ne - 2$

Hence x=-2 is an extraneous solution.