How do you solve $4x^2-8x>=0$ using a sign chart?

Question

1623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T17:56:02

The answer is $x in ]-oo, 0] uu [2, +oo [$

Let s factorise the expression

$4x^2-8x=4x(x-2)$

Let $f(x)=4x(x-2)$

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

Now we can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaa)$ $0$ $color(white)(aaaaa)$ $2$ $color(white)(aaaa)$ $+oo$

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

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

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

Therefore,

$f(x)>=0$ when $x in ]-oo, 0] uu [2, +oo [$

How do you solve 4x^2-8x>=04x^2-8x>=0 using a sign chart?