Question

How do you solve `2x^2=x+28`?

Answer 1

Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.

Explanation:

`2x^2=x+28`
`2x^2-x=28`
`2x^2-x-28=0`

Next, try to factor this expression.

`(2x+a)(x+b)=0`

Where `a xx b=-28` and `2bx + ax=-x` or `2b+a=-1`

The factors of -28 are:
`-(a xx b)` or `-(b xx a)`
`-(1 xx 28)`
`-(2 xx 14)`
`-(4 xx 7)`

Of these factors, only the last pair will allow you to make `2b+a=-1` by setting `b=-4` and `a=7`. But this means we can factor the expression above.

`(2x+a)(x+b)=0`
`(2x+7)(x-4)=0` which is only true when

`x=-7/2` or `x=4`.

Check your work by plugging these values back into the original expression.

Answer 2

`x_(1,2) = 1/4 +- 15/4`

Explanation:

You could solve this quadratic equation by completing the square.

To do that, start by getting your quadratic to the general form

`color(blue)(x^2 + b/ax = -c/a)`

by adding `-x` to both sides of the equation and dividing your terms by `2`

`2x^2 - x = color(red)(cancel(color(black)(x))) - color(red)(cancel(color(black)(x))) + 28`

`(color(red)(cancel(color(black)(2))) * x^2)/(color(red)(cancel(color(black)(2)))) - x/2 = 28/2`

`x^2 - x/2 = 14`

Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the square of a binomial.

The coefficient of the `x`-term will tell what term you need to add. More specifically, divide this coefficient by `2` and square the result to get

`(-1/2 * 1/2)^2 = 1/16`

Add `1/16` to both sides of the equation

`x^2 - x/2 + 1/16 = 14 + 1/16`

The left side of the equation can now be rewritten as

`x^2 - 2 * (1/4) * x + (1/4)^2 = (x-1/4)^2`

You now have

`(x-1/4)^2 = 225/16`

Take the square root of both sides of the equation

`sqrt( (x-1/4)^2) = sqrt(225/16)`

`x-1/4 = +- 15/4 => x_(1,2) = 1/4 +- 15/4`

The two solutions to the equation will thus be

`x_1 = 1/4 + 15/4 = color(green)(4)` and `x_2 = 1/4 - 15/4 = color(green)(-7/2)`

Answer 3

Solve `y = 2x^2 - x - 28 = 0` (1)

Explanation:

I use the new Transforming Method.
Transformed equation `y' = x^2 - x - 56 = 0`(2).
Factor pairs of (-56) --> (-4, 14)(-7, 8). This sum is 1 = -b. Two real roots of (2) are: -7 and 8.
The 2 real roots of equation (1) are: `(-7/2)` and `(8/2 = 4)`