Question

Solve$16y^2=32y^3+2y$

Short answer

The value is$y=\frac{1}{4}$

Step-by-step solution

Given information

Given the equation$16y^2=32y^3+2y$

Bring all terms to one side and solve

Here, we have

$$\begin{gathered} 16y^2=32y^3+2y \\ 32y^3+2y-16y^2=0 \\ 32y^3-16y^2+2y=0 \\ 2y(16y^2-8y+1)=0 \end{gathered}$$

Use zero factor property and calculate the value

Further solving,

$$\begin{gathered} 16y^2-8y+1=0 \\ 16y^2-4y-4y+1=0 \\ 4y(4y-1)-1(4y-1)=0 \\ (4y-1)(4y-1)=0 \\ {(4y-1)}^2=0 \\ \Rightarrow y=\frac{1}{4} \end{gathered}$$

Substitute and check the value

Substituting and checking, we get

$$\begin{gathered} 16{\left(\frac{1}{4}\right)}^2=32{\left(\frac{1}{4}\right)}^3+2\left(\frac{1}{4}\right) \\ 1=\frac{1}{2}+\frac{1}{2} \\ 1=1 \end{gathered}$$

So the value is correct