Question

Find real roots of question

Answer 1

Example: Find all real solutions of the equation x^1/3 +x^1/6 - 2=0?

We can simply use some simple “u-substitution.”

Let the dummy variable u = x^*(1/3). Then we can do the following:

u + u^(1/2) - 2 = 0

How I got u^(1/2) is by taking the square root of x^(1/3). Doing this gives us x^(1/6). (x^(1/3))^(1/2)— EVERYTHING BEING square rooted, which means we are multiplying the exponents together. (1/2)*(1/3) = (1/6).

u + u^(1/2) - 2 = 0.

Now u is really (u^(1/2))^2. When we square this, we have u^[(1/2)*2] = u^1.

So we can rewrite this as:

[u^(1/2)]^2 + u^(1/2) - 2 = 0

Use the quadratic formula to find the value of u:

u = (-b +/- sqrt(b^2 - 4ac))/2a

a = 1, b = 1, c = -2

u = (-1 +/- sqrt[1^2 - 4(1)(-2))/2(1)

u = (-1 +/- sqrt[1 + 8])/2

u = (-1 +/- sqrt(9))/2

u = (-1 +/- 3)/2

We have two solutions:

u = (-1 + 3)/2 and u = (-1 - 3)/2

u = -2/2 = -1 and u = (-4/2) = -2.

We are not done yet because we still have to find out the value of x. We let u=(x^(1/3)).

So this means we have x^(1/3) = -2 or x^(-1/3) = -1

Cubing both of these to get rid of the (1/3) exponent (also known as a cube root), we get:

x = (-1)^3 = -1 or x = (-2) = -8

So our solutions for this weird problem is x = -1, -8.

Hope this helps.

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