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

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Answer 1 · 2017-01-13T10:49:57

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

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 $p in ] -oo,3 ]uu [7,+ oo[$

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