Question

2x² +$\frac{x-1}{5}$ = 0

Answer 1

Given: 2x² +$\frac{x-1}{5}$ = 0
On rewriting this equation in standard form of the quadratic equation, we get:
10x² + x – 1 = 0
On comparing this equation with ax² + bx + c = 0, we get:
a = 10, b = 1, c = –1
For the quadratic equation ax² + bx + c = 0, we know that the quadratic formula is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On substituting a = 10, b = 1, c = –1, we get:

$$\begin{gathered} x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(10\right)\left(-1\right)}}{2\left(10\right)} \\ \Rightarrow x=\frac{-1\pm \sqrt{1+40}}{2\left(10\right)} \\ \Rightarrow x=\frac{-1\pm \sqrt{41}}{20} \\ \Rightarrow x=\frac{-1+\sqrt{41}}{20},\frac{-1-\sqrt{41}}{20} \\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}\text{is}x=\frac{-1+\sqrt{41}}{20},\frac{-1-\sqrt{41}}{20} \end{gathered}$$