Question

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x?

Answer 1

$\textbf{Concept used:}$

$\text{The roots of the quadratic equation ax^2+bx+c=0 are given by}$

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

$\text{Given equation can be written as}\bf\;6x^2-5x-3=0$

$\text{Here, a=6, b=-5, c=-3}$

$\text{Then}$

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

$x=\displaystyle\frac{5\pm\sqrt{25-4(6)(-3)}}{2(6)}$

$x=\displaystyle\frac{5\pm\sqrt{25+72}}{12}$

$x=\displaystyle\frac{5\pm\sqrt{97}}{12}$

$\therefore\textbf{The solution set is }\{\displaystyle\frac{5+\sqrt{97}}{12},\;\displaystyle\frac{5-\sqrt{97}}{12}\}$