irrational multiplication

[spacewater]

problem
\(\displaystyle 2x^{1/3}+3x^{2/3}=5\)

Where can I start to approach this type of problem? All I been doing was multiplication and the addition caught me off guard.

P.S
This question was on my test and I got partial credit due to my lucky guess (which was 1) plus my professor's generous partial credit policy.
Everything else was fine I did well on the test. Thanks to all the members who helped me to understand difficult problems. :mrgreen:

 

[Loren]

problem
\(\displaystyle 2x^{1/3}+3x^{2/3}=5\)

Let x[sup:1oizud2l]1/3[/sup:1oizud2l] = y.
Then x[sup:1oizud2l]2/3[/sup:1oizud2l] = y[sup:1oizud2l]2[/sup:1oizud2l].

Now, the equation becomes 2y + 3y[sup:1oizud2l]2[/sup:1oizud2l] = 5.

Solve that for y. When you have y, plug it back into your original "let" statements and solve for x.

 

[spacewater]

problem
\(\displaystyle 2x^{1/3}+3x^{2/3}=5\)

Let x[sup:2mox0syx]1/3[/sup:2mox0syx] = y.
Then x[sup:2mox0syx]2/3[/sup:2mox0syx] = y[sup:2mox0syx]2[/sup:2mox0syx].

Now, the equation becomes 2y + 3y[sup:2mox0syx]2[/sup:2mox0syx] = 5.

Solve that for y. When you have y, plug it back into your original "let" statements and solve for x.

so the only answer is 1

 

[stapel]

\(\displaystyle 2x^{1/3}+3x^{2/3}=5\)

Where can I start to approach this type of problem?

To learn the general technique for this "almost" quadratic, try here. Then:

. . . . .\(\displaystyle 2(x^{\frac{1}{3}})\, +\, 3(x^{\frac{1}{3}})^2\, =\, 5\)

. . . . .\(\displaystyle 3(x^{\frac{1}{3}})^2\, +\, 2(x^{\frac{1}{3}})\, -\, 5\, =\, 0\)

Since:

. . . . .\(\displaystyle 3y^2\, +\, 2y\, -\, 5\, =\, (3y\, +\, 5)(y\, -\, 1)\)

...how does your "quadratic" factor?

Once you have done the factorization, you can set the factors equal to zero, and then solve the resulting radical equations. :wink: