# math

posted by .

solve the equation. chech for extraneous solutions.
how do i do this.
1.x1/4=2
2.x2/3=16
3.x1/2=8

• math -

Use the rules of exponents, for example:
xa+b=xa+xb
xa*b=(xa)^b

x1/4=2
raise to the fourth power on each side:
(x1/4)^4=2^4
x(1/4)*4=16
x=16

However, note that since 2^4=16, (-2)^4=16, the roots to the equation appear to be 2,-2.

Since we have squared both sides of the equation, twice, we need to check for extraneous roots (that may have been introduced because of squaring) by substituting the possible answers into the original equation.

We find that -2 is not a correct answer, thus an extraneous root.

I will leave you to tackle the remaining questions.

## Similar Questions

1. ### algebra

Solve each equation. Watch for extraneous solutions 5/y-3 = y+7/2y-6 +1
2. ### Math

Solve the equation. Identify any extraneous solutions. x = sqrt root 3x + 40
3. ### Alg 2

Solve the equation. Check for extraneous solutions. 10 ln 100x - 3 = 117 Thank you in advanced
4. ### Alg 2

Solve the equation. Check for extraneous solutions. ln x + ln(x-2) = 1
5. ### math

Solve the equation check for extraneous solutions 6/6-4x/=8x+4 I used/ / for the brackets posssible answers x=1 x=5/2 or x=2 x=5/2 x=5/2 or x=1 Not looking for just the answer, would like to know how to get it.
6. ### algebra

Solve the equation check for extraneous solutions 9[9-8x]=2x+3 please explain
7. ### Precalculus

Solve (log x)^2 = log x Also, when will their be a possibility that an equation will have an extraneous solution?
8. ### Math-Clarification

Identify any real and extraneous solutions for the equations below. Sqrt p = -1 So, would this have no real or extraneous solutions?
9. ### algebra 2

Solve the equation. Check for extraneous solutions. 9|9-8x|=2x+3 can anyone please show me how to do this problem?
10. ### Math

1) Solve the equation for p. Identify any extraneous solutions. √p=-1 A) -1 is a solution of the original equation. 1 is an extraneous solution. B) 1 is a solution of the original equation. C) 1 is a solution of the original …

More Similar Questions