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

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

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Answer 1 · 2016-11-26T18:42:34

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

Explanation:

Let's rewrite the equation as

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

Let's do a sign chart

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

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

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

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

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

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

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

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