Question

Find the real solutions, if any, of the following equation. Use the quadratic formula.

Answer 1

EXPLANATION

Given the equation 6x^2 = 5x

We can apply the following procedure :

Subtracting -5x to both sides:

$6x^2-5x=0$

Apply exponent rule:

$a^{(b+c)}=a^ba^c$$x^2=\times$$=6\times-5x$

Factor out common term x:

$x\mleft(6x-5\mright)=0$$\mathrm{Using\: the\: Zero\: Factor\: Principle\colon\quad \: If}\: ab=0\: \mathrm{then}\: a=0\: \mathrm{or}\: b=0$$x=0\quad \mathrm{or}\quad \: 6x-5=0$$\mathrm{Add\: }5\mathrm{\: to\: both\: sides}\text{ from 6x -5=0}$$6x-5+5=0+5$$Simplify\colon$$6x=5$$\mathrm{Divide\: both\: sides\: by\: }6$$\frac{6x}{6}=\frac{5}{6}$$Simplify\colon$$x=\frac{5}{6}$$\mathrm{The\: solutions\: to\: the\: quadratic\: equation\: are\colon}$$x=0,\: x=\frac{5}{6}$

Hence, the solution set is as follows:

{0 , 5/6}