Question

How to solve -3x² + 5 < 2x - 1

Answer 1

−3x² + 5 = 2x −1

−3x² + 5−(2x −1) = 2x −1 −(2x −1) (Subtract 2x-1 from both sides)

−3x²−2x + 6 = 0

For this equation: a=-3, b=-2, c=6

−3x²+−2x+6=0

===> (Use quadratic formula with a=-3, b=-2, c=6)

`\large\displaystyle\text{$\begin{gathered}\sf x=\frac{-b\pm\sqrt{b^(2)-4ac } }{2a} \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf x=\frac{-(-2)\pm\sqrt{(-2)^(2)-4(-3)(6) } }{2(-3)} \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf x=(-2\pm√(76) )/(-6 ) \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf x=(-1)/(3)+(-1)/(3)√(19) \ or \ x=(-1)/(3)+(1)/(3)√(19) \end{gathered}$}`

Check intervals in between critical points. (Test values in the intervals to see if they work.)

`\large\displaystyle\text{$\begin{gathered}\sf x < (-1)/(3)+(-1)/(3)√(19) \ \ (Works\:in\:original\:inequality) \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf (-1)/(3)+(-1)/(3)√(19) < x < (-1)/(3)+(1)/(3)√(19) \ \ (Doesn't\:work\:in\:original\:inequallity) \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf x > (-1)/(3)+(-1)/(3)√(19) \ \ (Works\:in\:original\:inequality) \end{gathered}$}`

`\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{x < (-1)/(3)+(-1)/(3)√(19) \ \ or \ x > (-1)/(3)+(1)/(3)√(19) } \end{gathered}$}}}`

Michael Spymore