Question

How do you solve for `x` in this problem? `4x^2 - 3 = -x`?

Answer 1

`x=-1" or "x=3/4`

Explanation:

`"rearrange into standard form "color(white)(x)ax^2+bx+c=0`

`"add "x" to both sides"`

`rArr4x^2+x-3=0larrcolor(blue)"in standard form"`

`"factor the quadratic using the a-c method"`

`"the factors of the product "4xx-3=-12`

`"which sum to + 1 are + 4 and - 3"`

`"split the middle term using these factors"`

`4x^2+4x-3x-3=0larrcolor(blue)"factor by grouping"`

`color(red)(4x)(x+1)color(red)(-3)(x+1)=0`

`"take out the "color(blue)"common factor "(x+1)`

`rArr(x+1)(color(red)(4x-3))=0`

`"equate each factor to zero and solve for x"`

`x+1=0rArrx=-1`

`4x-3=0rArrx=3/4`

Answer 2

Use the quadratic formula

Explanation:

First we rearrange;

`4x^2 -3 = -x`
`4x^2 +x-3=0`

now we have `ax^2 + bx + c =0` format, so we can use our quadratic equation to solve.

`x = (-b+sqrt(b^2-4ac))/(2a)`
and
`x = (-b-sqrt(b^2-4ac))/(2a)`
After subbing everything in we get `x=-1` or `x=.75`

Answer 3

the answer
`x=-1 and x=3/4`

Explanation:

show below

`4x^2 - 3 = -x`

`4x^2+x-3=0`

`(x+1)*(4x-3)=0`

`x+1=0 => x=-1`

`4x-3=0 => 4x=3 => x=3/4`