Question

Solve the given quadratic inequality. Write the answer in interval notation.

Answer 1

`(-\infty,\text{ -0.6458})\cup(4.6458,\text{ }\infty)`

Step-by-step explanation:

`\begin{gathered} x^2\text{ - 4x > 3} \\ x^2\text{ - 4x - 3 > 0} \end{gathered}`
`\begin{gathered} \text{using almighty formula:} \\ x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x\text{ > }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\text{ (for the inequality)} \\ a\text{ = 1, b = -4, c = -3} \\ x\text{ > }\frac{-(-4)_{}\pm\sqrt[]{(-4)^2-4(1)(-3)}}{2(1)} \\ x\text{ > }\frac{4_{}\pm\sqrt[]{16+12}}{2} \end{gathered}`
`\begin{gathered} x\text{ > }\frac{4_{}\pm\sqrt[]{28}}{2}\text{ } \\ x\text{ > }\frac{4_{}\pm\sqrt[]{7*4}}{2} \\ x\text{ > }\frac{4_{}\pm2\sqrt[]{7}}{2} \\ x\text{ > }\frac{2(2_{}\pm\sqrt[]{7})}{2} \\ x\text{ > }\frac{2_{}\pm\sqrt[]{7}}{2} \\ x\text{ > }2_{}+\sqrt[]{7}\text{ or }x\text{ < }2_{}-\sqrt[]{7} \\ x\text{ > }4.6458\text{ or }x\text{ < }-0.6458\text{ } \\ \end{gathered}`
`\begin{gathered} \text{Interval notation:} \\ (-\infty,\text{ -0.6458})\cup(4.6458,\text{ }\infty) \end{gathered}`