How do you solve $x^2+21>10x$ using a sign chart?

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Answer 1 · 2016-12-30T19:09:30

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

Let's rearrange the equation

$x^2-10x+21>0$

Let $f(x)=x^2-10x+21$

The domain of $f(x)$ is $D_f(x) = RR$

Let's factorise

$f(x)=(x-3)(x-7)$

Now, we can establish the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $3$ $color(white)(aaaaa)$ $7$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-3$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-7$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)>0$ when $x in ] -oo,3 [ uu ] 7, +oo[$

How do you solve x^2+21>10xx^2+21>10x using a sign chart?