How do you solve $x^2-10<3x$ ?

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Answer 1 · 2016-11-26T18:42:34

THe answer is $x in ] -2,5 [$

Let's rewrite the equation as

$f(x)=x^2-3x-10=(x+2)(x-5)<0$

Let's do a sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2$ $color(white)(aaaa)$ $5$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+2$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

So, $f(x)<0$ for $x in ] -2,5 [$

graph{x^2-3x-10 [-25.66, 25.65, -12.83, 12.85]}

How do you solve x^2-10<3xx^2-10<3x?