Question

How do you solve `112v ^ { 2} - 360v = - 200`?

Answer 1

`v=5/2 and 5/7`

Explanation:

Assuming you can not spot how to deal with the numbers. Then try this sort of approach.

Bit of a long way round but we get there.

Given: `112v^2-360v=-200`

divide everything by 10

`11.2v^2-36v=-20`

Get rid of the decimal by multiplying by 5 `(0.2xx5=1)`

`56v^2-180v=-100`

halve everything

`28v^2-90v=-50`

halve again

`14v^2-45v=-25`

Add 25 to both sides

`14v^2-45v+25=0`

Consider some of the factors (first guess)

`14=2xx7` giving the structure: `(2v+?)(7v+?)`

`25=5xx5` and I notice that `5xx7=35` and `2xx5=10` where `35+10=45`

So lets test `color(green)(color(blue)((2v-5))(7v-5))=0`

.......................................................................................
Multiply everything in the right brackets by everything in the left.

`color(green)(color(blue)(2v)(7v-5)color(blue)(" "+" "5)(7v-5))=0`
`14v^2-10v" "+" "-35v+25=0`

`14v^2-45v+25 =0color(red)(" "larr" Works")`
.....................................................................................

So it is correct to state that:

`(2v-5)(7v-5)=0`

Set `2v-5=0" "=>" "v=+5/2`

Set `7v-5=0" "=>" "v=+5/7`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check") " " 112v^2-360v=-200`

set `v=5/2`

`112(5/2)^2-360(5/2) -> -200` as required

set `v=5/7`

`112(5/7)^2-360(5/7) -> -200` as required

Answer 2

`v =5/2" or " v= 7/2`

Explanation:

The first thing to notice is that we are working with a quadratic trinomial. Make it equal to zero.

`112v^2 -360v +200 =0" "larr` divide by the obvious factor of `4`.

`28v^2 -90v +50=0" "larr` divide by `2`

`14v^2 -45v +25=0`

Find factors of `14 and 25` which add to give 45.

`" "14 and 25` find factors and cross multiply
`" "2" "5 " "rarr 7 xx 5 = 35`
`" "7" "5 " "rarr 2 xx 5 = ul10`
`color(white)(xxxxxxxxxxxxxxxxxxx)45`

These are the correct factors, the signs are both negative.

`(2v-5)(7v-5) =0`

Set each factor equal to `0` and solve.

`2v-5=0 " "rarr v =5/2`

`7v -5=0" "rarr v = 5/7`