Question

How do you solve `-p^2+10p-7<=14`?

Answer 1

The answer is `p in ] -oo,3 ]uu [7,+ oo[`

Explanation:

Let's rewrite the equation

`-p^2+10p-7<=14`, `hArr`, `p^2-10p+21>=0`

Let 's factorise the LHS

`(p-3)(p-7)>=0`

Let `f(p)=(p-3)(p-7)`

Now we can make the sign chart

`color(white)(aaaa)` `p` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `3` `color(white)(aaaaa)` `7` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `p-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `p-7` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(p)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(p)>=`, when `p in ] -oo,3 ]uu [7,+ oo[`