Question

How do you solve `32x^2-18=0`?

Answer 1

See the solution process below:

Explanation:

First, add `color(red)(18)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`32x^2 - 18 + color(red)(18) = 0 + color(red)(18)`

`32x^2 - 0 = 18`

`32x^2 = 18`

Next, divide each side of the equation by `color(red)(32)` to isolate the `x^2` term while keeping the equation balanced:

`(32x^2)/color(red)(32) = 18/color(red)(32)`

`(color(red)(cancel(color(black)(32)))x^2)/cancel(color(red)(32)) = (2 xx 9)/color(red)(2 xx 16)`

`x^2 = (color(red)(cancel(color(black)(2))) xx 9)/color(red)(color(black)(cancel(color(red)(2))) xx 16)`

`x^2 = 9/16`

Now, take the square root of each side of the equation to solve for `x` while keeping the equation balanced. Remember, taking the square root of a number produces a negative and positive result:

`sqrt(x^2) = sqrt(9/16)`

`x = sqrt(9)/sqrt(16)`

`x = +-3/4`

Answer 2

x = 0.75

Explanation:

`32x^2-18=0`
`32x^2-18+18=0+18`
`32x^2=18`
`32x^2/32=18/32`
`x^2=0.5625`
`sqrt(x^2)=sqrt(0.5625)`
`x=0.75`

Answer 3

`color(blue)(x=3/4` or `color(blue)(x=-3/4`

Explanation:

`32x^2-18=0`

Take out the common factor first

`:.2(16x^2-9)=0`

`:.2(4^2x^2-3^2)=0`

Divide both sides by `2`

`:.(4^2x^2-3^2)=0`

`:.(4x-3)(4x+3)=0`

`:.4x-3=0,4x+3=0`

`:.4x=3,4x=-3`

`:.color(blue)(x=3/4,x=-3/4`

substitute `color(blue)(x=3/4=0.75`

`:.32(color(blue)0.75)^2-18=0`

`:.32(0.5625)-18=0`

`18-18=0`

substitute `x=-3/4=-0.75`

`:.32(color(blue)-0.75)^2-18=0`

`:.32(0.5625)-18=0`

`:.18-18=0`