Question

How do you solve `6x ^ { 2} - 5= 4`?

Answer 1

`x=sqrt(3/2)` or `x=-sqrt(3/2)`

Explanation:

First we subtract `4` from both sides.
`6x^2-5=4`
`6x^2-5color(blue)-color(blue)4=4color(blue)-color(blue)4`
`6x^2-9=0`

Next we add `5` to both sides of the equation:
`6x^2-5color(blue)+color(blue)5=4color(blue)+color(blue)5->`
`6x^2=9`

Divide both sides by `6`:
`(6x^2)/6=9/6->`

`x^2=9/6->`

`x^2=3/2`

We will solve `x^2=3/2` using this:If `x^2=y`, then `x=sqrty` or `-sqrty`

Therefore, `x=sqrt(3/2)` or `x=-sqrt(3/2)`

Just for fun: `x=1.2247`... `x=1.2247...`

Answer 2

Solution : `x =sqrt (3/2) , x = - sqrt (3/2)`

Explanation:

`6x^2-5=4 or 6x^2=5+4 or 6x^2=9` or

`x^2=9/6 or x^2 = 3/2 or x = +- sqrt (3/2)`

Solution : `x =sqrt (3/2) , x = - sqrt (3/2)` [Ans]

Answer 3

`x=+-sqrt(3/2)`

Explanation:

`"isolate the "x^2" term by manipulations"`

`"add 5 to both sides"`

`6x^2cancel(-5)cancel(+5)=4+5`

`rArr6x^2=9`

`"divide both sides by 6"`

`(cancel(6)x^2 )/cancel(6)=9/6=3/2`

`rArrx^2=3/2`

`color(blue)"take the square root of both sides"`

`sqrt(x^2)=+-sqrt(3/2)larr" note plus or minus"`

`rArrx=+-sqrt(3/2)larr" exact solutions"`