Question

How do you solve `6x^2 - 2 = x`?

Answer 1

`x_1=2/3`
`x_2=-1/2`

Explanation:

From the given equation `6x^2-2=x`

Transpose the x term to the left of the equation

`6x^2-2=x`

`6x^2-x-2=0`

We can solve this by factoring method

`(3x - 2)(2x + 1 )=0`

After factoring, equate each factor to 0
First factor equate to 0
`3x-2=0`

`3x=2`

`(3x)/3=2/3`

`(cancel3x)/cancel3=2/3`

`x=2/3`

Second factor equate to 0
`2x+1=0`
`2x=-1`

`(2x)/2=(-1)/2`

`(cancel2x)/cancel2=(-1)/2`

`x=-1/2`

God bless....I hope the explanation is useful.

Answer 2

`x=2/3,-1/2`

Explanation:

`color(blue)(6x^2-2=x`

Subtract `x` both sides

`rarr6x^2-2-x=cancel(x-x`

`rarr6x^2-2-x=0`

Rewrite in standard form

`rarrcolor(purple)(6x^2-x-2=0`

Now,this is a Quadratic equation (in form `ax^2+bx+c=0`)

Use Quadratic formula

`color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)`

Remember that `a,` `bandc` are the coeficients of `x^2,` `xand -2`

Where

`color(red)(a=6,b=-1,c=-2`

`rarrx=(-(-1)+-sqrt(-1^2-4(6)(-2)))/(2(6))`

`rarrx=(1+-sqrt(1-4(6)(-2)))/(12)`

`rarrx=(1+-sqrt(1-(-48)))/(12)`

`rarrx=(1+-sqrt(1+48))/(12)`

`rarrx=(1+-sqrt(49))/(12)`

`color(orange)(rarrx=(1+-7)/(12)`

Now we have `2` solutions

`color(violet)(x=(1+7)/(12)=8/12=2/3`

`color(indigo)(x=(1-7)/(12)=-6/12=-1/2`

`color(blue)( :.ul bar |x=2/3,-1/2|`