How do you solve $- 2 x^{2} + 10 < 8 x$ $-2x^2+10<8x$ using a sign chart?

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How do you solve $- 2 x^{2} + 10 < 8 x$ $-2x^2+10<8x$ using a sign chart?

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Answer 1 · 2016-10-29T14:16:25

The solution are $x < - 5$ $x<-5$ and $x > 1$ $x>1$

Explanation:

Let's rewrite the expression
$f (x) = 2 x^{2} + 8 x - 10 > 0$ $f(x)=2x^2+8x-10>0$
so factorising
$(2 x - 2) (x + 5) = 2 (x - 1) (x + 5) > 0$ $(2x-2)(x+5)=2(x-1)(x+5)>0$
So the values of x to be taken in consideration are $x = 1$ $x=1$ and $x = - 5$ $x=-5$

Let 's do the sign chart
$x$ $x$ $a a a$ $color(white)(aaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 5$ $-5$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$
$x - 1$ $x-1$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$
$x + 5$ $x+5$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$
$f (x)$ $f(x)$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

So the values of x are $x < - 5$ $x<-5$ and $x > 1$ $x>1$

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How do you solve $- 2 x^{2} + 10 < 8 x$ $-2x^2+10<8x$ using a sign chart?

1 Answer

Explanation:

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How do you solve $- 2 x^{2} + 10 < 8 x$ $-2x^2+10<8x$ using a sign chart?