Question

How do you solve `5v ^ { 2} - 40= - 10v`?

Answer 1

`v= 2`
`v= -4`

Explanation:

First, make sure the equation is in the standard form
`ax^2 + bx +c = 0`

`5v^2 + 10v - 40 = 0`

If `a!= 1`, the `ac` method is applied.
In this case, `a= 5` so we have to use the ac method.

What is the ac method???

The ac method is simply the value of `a` times the value of `c`

`a= 5`
`b= 10`
`c= -40`

`a*c`= -200

Now we have to find the numbers that will give a product of -200 as well as a sum of 10.

`-10 * 20 = -200 (ac)`
`-10 + 20 = 10 (b)`

Bingo! Now that we have these numbers, we rewrite the equation substituting -10 and 20 for b.

5`v^2` - 10v + 20v -40 = 0 ...this is the same as the inital equation when simplified.

Now, we factor by grouping and pulling out the gcf.

`(5v^2 - 10v) + (20v - 40) = 0`

`5v(v - 2) + 20(v - 2) = 0`

`(5v + 20)(v - 2) = 0`

Solve for v
`5v + 20 = 0`
`5v = -20`

Dividing both sides by 5, we get
`(cancel(5)v)/cancel5 = (-20)/5`
`v = -4`

Solve the other also for v
`v - 2 = 0`
`v = 2`