Asked by Anonymous
solve the equation. chech for extraneous solutions.
how do i do this.
1.x1/4=2
2.x2/3=16
3.x1/2=8
how do i do this.
1.x1/4=2
2.x2/3=16
3.x1/2=8
Answers
Answered by
MathMate
Use the rules of exponents, for example:
x<sup>a+b</sup>=x<sup>a</sup>+x<sup>b</sup>
x<sup>a*b</sup>=(x<sup>a</sup>)^b
x<sup>1/4</sup>=2
raise to the fourth power on each side:
(x<sup>1/4</sup>)^4=2^4
x<sup>(1/4)*4</sup>=16
x=16
However, note that since 2^4=16, (-2)^4=16, the roots to the equation <i>appear</i> 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 <i>possible</i> answers into the original equation.
We find that -2 is not a correct answer, thus an extraneous root.
The answer is therefore 2.
I will leave you to tackle the remaining questions.
x<sup>a+b</sup>=x<sup>a</sup>+x<sup>b</sup>
x<sup>a*b</sup>=(x<sup>a</sup>)^b
x<sup>1/4</sup>=2
raise to the fourth power on each side:
(x<sup>1/4</sup>)^4=2^4
x<sup>(1/4)*4</sup>=16
x=16
However, note that since 2^4=16, (-2)^4=16, the roots to the equation <i>appear</i> 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 <i>possible</i> answers into the original equation.
We find that -2 is not a correct answer, thus an extraneous root.
The answer is therefore 2.
I will leave you to tackle the remaining questions.