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# 1/3(x+1)^(-2/3)=0

### 2 Answers by Expert Tutors

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (276 lesson ratings) (276)
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There is one missing detail here. It's true that x = -1 is the only candidate for the solution and would have multiplicity 2, however, you must plug it into the original equation to see if it works. It turns out that it doesn't because x+1=0 is raised to the negative power and 0p is undefined when p < 0.

Indeed, the graph of y = 1/3(x + 1)-2/3 has a point-discontinuity at the point (-1,0).

Shefali J. | A Complete Math Tuition SolutionA Complete Math Tuition Solution
4.9 4.9 (263 lesson ratings) (263)
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Hello Diana,

The problem is

1/3(x + 1)(-2/3) = 0

Since there's a negative exponent in the denominator so, bring it to the numerator.

[(x + 1)(2/3)] / 3 = 0   (as 1/x-m = xm)

Multiply by 3 on both sides to get rid of 3 in the denominator on the left hand side. Now, it'll look like this

3 * [(x + 1)(2/3)] / 3 = 0 * 3 (cancel the 3 on the left side)

Finally, you'll get

[(x + 1)(2/3)]  = 0

Take cube on both sides of the equation

[(x + 1)(2/3)]3= (0)3

(x + 1)(2/3)*3 = 0    (as  (xm)n = x m.n)

Now it'll look like this

(x + 1)2 = 0 or (x + 1)(x + 1) = 0

x + 1 = 0 and x + 1 = 0

so x has two values

x = -1 and x = -1 (answer)

Hope this helps.