Question

Solve 2x^2+x-4=0x^2+1/2x+___=2+____

Answer 1

We have the following expression:

$2x^2+x-4=0$

By dividing both sides by 2, we get

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

At this point we can apply the quadratic formula:

$\begin{gathered} \text{For ax}^2+bx+c=0 \\ x\text{ is given by} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

where a is 1, b is 1/2 and c is -2. By substituting these values, we get

$x=\frac{-\frac{1}{2}\pm\sqrt[]{(\frac{1}{2})^2-4(1)(-2)}}{2}$

which gives

$\begin{gathered} x=\frac{-\frac{1}{2}\pm\sqrt[]{\frac{1}{4}+8}}{2} \\ x=\frac{-\frac{1}{2}\pm\sqrt[]{\frac{33}{4}}}{2} \\ x=\frac{-\frac{1}{2}\pm\frac{\sqrt[]{33}}{2}}{2} \\ x=-\frac{1}{4}\pm\frac{\sqrt[]{33}}{4} \end{gathered}$

Then, the solutions are

$\begin{gathered} x=\frac{-1+\sqrt[]{33}}{4} \\ \text{and} \\ x=\frac{-1-\sqrt[]{33}}{4} \end{gathered}$