Question

Solve each equation.
$y^{\frac{2}{3}}+2y^{\frac{1}{3}}-3=0$

Short answer

The acquired solutions are:

$y=1,y=-27$

But the only solution that will give a real value is:

$y=1$

Step-by-step solution

Step 1. Given information.

We have:

$y^{\frac{2}{3}}+2y^{\frac{1}{3}}-3=0$

Step 2. Rewrite the equation as standard form  

Use the following exponent property: $a^{\frac{n}{m}}={\left(a^{\frac{1}{m}}\right)}^n$

$$\begin{gathered} y^{\frac{2}{3}}={\left(y^{\frac{1}{3}}\right)}^2 \\ {\left(y^{\frac{1}{3}}\right)}^2+2y^{\frac{1}{3}}-3=0 \\ \mathrm{Rewrite}\hspace{0.17em}\mathrm{the}\hspace{0.17em}\mathrm{equation}\hspace{0.17em}\mathrm{with}\hspace{0.17em}{y}^{\frac{1}{3}}=u \\ {u}^2+2u-3=0 \end{gathered}$$

Step 3. Solve the quadratic equation. 

We have:

$u^2+2u-3=0$

For a quadratic equation of the form $a{x}^2+bx+c=0$the solutions are:

$$\begin{gathered} x_{1,\hspace{0.17em}2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For}\hspace{0.17em}a=1,\hspace{0.17em}b=2,\hspace{0.17em}c=-3 \\ {u}_{1,\hspace{0.17em}2}=\frac{-2\pm \sqrt{2^2-4·\hspace{0.17em}1·\left(-3\right)}}{2·\hspace{0.17em}1} \\ {u}_{1,\hspace{0.17em}2}=\frac{-2\pm \hspace{0.17em}4}{2·\hspace{0.17em}1} \\ u=1,\hspace{0.17em}u=-3 \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}\hspace{0.17em}\mathrm{back}\hspace{0.17em}u=y^{\frac{1}{3}},\hspace{0.17em}\mathrm{solve}\hspace{0.17em}\mathrm{for}\hspace{0.17em}y \\ {y}^{\frac{1}{3}}=1 \\ y=1 \\ {y}^{\frac{1}{3}}=-3 \\ y={(-3)}^3=-27 \end{gathered}$$

Step 4. Verify the solutions.

At $y=1:$

localid="1667913904433" $$\begin{gathered} {\left(1\right)}^{\frac{2}{3}}+2{\left(1\right)}^{\frac{1}{3}}-3=0 \\ 1+2-3=0 \end{gathered}$$

$0=0$

Hence, verified.

At $y=-3:$

$$\begin{gathered} {(-3)}^{\frac{2}{3}}+2{(-3)}^{\frac{1}{3}}-3=0 \\ {\left(-1\right)}^{\frac{2}{3}}·\hspace{0.17em}{3}^{\frac{2}{3}}+2{\left(-1\right)}^{\frac{1}{3}}·\hspace{0.17em}{3}^{\frac{1}{3}}-3=0 \end{gathered}$$

Both sides are not equal.

Hence, not verified.

Therefore, localid="1667914096014" $y=1$is the real solution.