Question

Can someone solve this equation using the quadratic formula and simplifying in radical form if needed

Answer 1

For a quadratic equation of the form:

`av^2+bv+c=0`

The quadratic formula is:

`v_(1,2)=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a}`

In this case, we have the eqaution:

`11v^2+8v=4`

First, let's rest 4 on both sides to get 0 in the right hand side:

`11v^2+8v-4=0`

Then we can use the quadratic formula:

`v_(1,2)=\frac{-8\pm\sqrt[]{8^2-4\cdot11\cdot(-4)}}{2\cdot11}`

And solve:

`\begin{gathered} v_(1,2)=\frac{-8\pm\sqrt[]{64^{}+176}}{22} \\ v_(1,2)=\frac{-8\pm\sqrt[]{240}}{22} \\ v_(1,2)=\frac{-8\pm\sqrt[]{16}\sqrt[]{15}}{22} \\ v_(1,2)=\frac{-8\pm4\sqrt[]{15}}{22} \\ v_(1,2)=\frac{-4\pm2\sqrt[]{15}}{11} \end{gathered}`

Then the two solutions are:

`\begin{gathered} v_1=\frac{-4-2\sqrt[]{15}}{11} \\ v_2=\frac{-4+2\sqrt[]{15}}{11} \end{gathered}`