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=*`
`=6*-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`
`(6x)/(6)=(5)/(6)`
`Simplify\colon`
`x=(5)/(6)`
`\mathrm{The\: solutions\: to\: the\: quadratic\: equation\: are\colon}`
`x=0,\: x=(5)/(6)`

Hence, the solution set is as follows:

{0 , 5/6}