Question

Solve the equation using the quadratic formula. 2y^2 + 4y + 1 = 0

Answer 1

The solution to the quadratic equation is;

`\begin{gathered} y=-1+\frac{\sqrt[]{2}}{2} \\ \text{and} \\ y=-1-\frac{\sqrt[]{2}}{2} \\ y=-1+\frac{\sqrt[]{2}}{2},-1-\frac{\sqrt[]{2}}{2} \end{gathered}`

Step-by-step explanation:

Given the quadratic equation;

`2y^2+4y+1=0`

Applying quadratic formula;

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Substituting the coefficients of the quadratic equation;

`\begin{gathered} y=\frac{-4\pm\sqrt[]{4^2-4(2)(1)}}{2(2)} \\ y=\frac{-4\pm\sqrt[]{16^{}-8}}{4} \\ y=\frac{-4\pm\sqrt[]{8}}{4} \\ y=\frac{-4\pm2\sqrt[]{2}}{4} \\ y=\frac{-2\pm\sqrt[]{2}}{2} \end{gathered}`

Therefore, the solution to the quadratic equation is;

`\begin{gathered} y=-1+\frac{\sqrt[]{2}}{2} \\ \text{and} \\ y=-1-\frac{\sqrt[]{2}}{2} \\ y=-1+\frac{\sqrt[]{2}}{2},-1-\frac{\sqrt[]{2}}{2} \end{gathered}`