Question

Question #ae88d

Answer 1

See below.

Explanation:

If the question is stated correctly as `5x^2+4=x`:

First, we move the `x` to the left-hand side by subtracting `x` from both sides to make the right-hand side equal `0`:
`5x^2-x+4=0`

Since this cannot be easily factored, we use the quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`
where `a=5,b=-1,c=4` according to the given equation
So, by substitution, we have:
`x=(1+-sqrt((-1)^2-4(5)(4)))/(2(5))`
`x=(1+-sqrt(-79))/10`
`x=(1+-isqrt(79))/10`

Since we cannot simplify this further, our solutions are:
`x=(1+isqrt(79))/10,(1-isqrt(79))/10`

If the question was intended to be stated as `5x^2-4=x`:

We follow a similar procedure:
We move the `x` to the left-hand side by subtracting `x` from both sides to make the right-hand side equal `0`:
`5x^2-x-4=0`

This can be easily factored as:
`(5x+4)(x-1)=0`
So we get that `5x+4=0` or `x-1=0` and solve for `x`:
`5x+4=0`
`x=-4/5`

`x-1=0`
`x=1`

So our solutions for this question are: `x=-4/5,1`