Question

4x^2+x-5=-6x to the nearest tenth.

Answer 1

SOLUTION

We want to solve

$\begin{gathered} 4x^2+x-5=-6x \\ bringing\text{ 6x to the other side, we have } \\ 4x^2+x+6x-5=0 \\ 4x^2+7x-5=0 \\ This\text{ is in the form of the quatric equation } \\ ax^2+bx+c=0 \\ So,\text{ it means } \\ a=4,b=7,c=-5 \end{gathered}$

Using the quadratic formula

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

We have

$\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-7\pm\sqrt{(-7)^2-4\times4\times-5}}{2\times4} \\ x=\frac{-7\pm\sqrt{49+80}}{8} \\ x=\frac{-7\pm\sqrt{129}}{8} \end{gathered}$

So, either

$\begin{gathered} x=\frac{-7+\sqrt{129}}{8} \\ x=\frac{-7+11.35781669}{8} \\ x=0.544727 \\ x=0.5 \end{gathered}$

Or

$\begin{gathered} x=\frac{-7-\sqrt{129}}{8} \\ x=\frac{-7-11.35781669}{8} \\ x=-2.294727 \\ x=-2.3 \end{gathered}$

Hence the answer is x = 0.5 or -2.3 to the nearest tenth