Q44. - Algebra 2

Question

Solve each equation.

 

 $y^4+4y^3+4y^2=0$

Step-by-Step Solution

Answer

The solutions are -2 and 0.

Step 1. Given Information.

Given, to solve the equation  $y^4+4y^3+4y^2=0$.

Step 2. Calculation .

Solving the equation:

 

 $\begin{array}{l}{\left(y^2\right)}^2+2y^2\cdot \left(2y\right)+{\left(2y\right)}^2=0\\ {\left(y^2\right)}^2+y^2\cdot \left(2y\right)+y^2\cdot \left(2y\right)+{\left(2y\right)}^2=0\left(\text{Writing it in the quadratic form}\right)\\ \left(y^2\right)\left(y^2+2y\right)+\left(2y\right)\left(y^2+2y\right)=0\\ \left(y^2+2y\right)\left(y^2+2y\right)=0\left(\text{Factoring the trinomial}\right)\\ {\left(y^2+2y\right)}^2=0\\ y^2+2y=0\\ y\left(y+2\right)=0\end{array}$

 

Using the Zero Product Property:

 $\begin{array}{l}y=0\text{ or }y+2=0\\ y=0\text{ or }y=-2\end{array}$

 

Step 3. Conclusion .

Hence, the solutions are -2 and 0.