Question

How do you solve `2x ^ { 2} + 57x = 497`?

Answer 1

`x = 7` , `x=-35.5`

Explanation:

First we must convert our initial equation so that it fits satisfies `ax^2 + bx + c = 0`.

`2x^2 + 57x = 497` becomes `2x^2 + 57x - 497 = 0`.

From here we conclude the following:

`a=2`
`b=57`
`c=-497`

Now we use the quadratic formula to find the value of `x`.

`x = (-b +- sqrt(b^2 -4ac))/(2a)`

`x = (-57 +- sqrt(57^2 -4(2)(-497)))/(2(2))`

`x = (-57 +- sqrt(3249 -8(-497)))/(4)`

`x = (-57 +- sqrt(3249 + 3976))/(4)`

`x = (-57 +- sqrt(7225))/(4)`

`x = (-57 +- 85)/(4)`

`x = (-57 + 85)/4` , `x=(-57-85)/4`

`x = (28)/(4)` , `x = (-142)/(4)`

`x = 7` , `x=-35.5`

In this case it is not specified whether we want the positive `x` value, so we are left with two solutions which are `x = 7` and `x=-35.5`.

Answer 2

Given: `2x^2+57X=497`

We can quickly turn this into an easily recognizable quadratic equation by subtracting 497 from both sides to get:

`2x^2+57X-497=0`

There are a couple of "give aways" we can see on this equation that will take us to the solution:

1) The `x` exponent is `^2` so there are two answers for `x` and there will need to be two brackets in the solution.

2) the multiplier `497` looks like it should be divided by `7`.

`497/7=71`

Setting up the brackets: `(2x ...71)(x ...7)=0`

There is a `+ and -` in the given equation so we know there will be a `+ and -` in the brackets. We also need the larger value to be `+`, so that will be the `71x`:

`(2x + 71)(x - 7)=0` and we see that `71x - 7xx2x = 57x`

Then: `(2x + 71)=0`

`2x = - 71`

`x = - 71/2`

And: `(x-7) = 0`

`x = 7`

To check, put the answers back into the `given` equation:

`2x^2+57x=497`

`2(7)^2+57(7)=497`

`2(49)+399=497`

`98+399=497`

And the other factor:

`2x^2+57x=497`

`2(-71/2)^2+57(-71/2)=497`

`(5041/2) + (-4047/2) = 497`

`5041 - 4047 = 2xx497`

`994=994`