How do you solve $2x^2-6x+3>=0$ using a sign chart?

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Answer 1 · 2017-06-28T06:16:22

The solution is $x in (-oo,(3-sqrt3)/2] uu [(3+sqrt3)/2,+oo)$

We need the roots of the equation

$2x^2-6x+3=0$

The discriminant is

$Delta=b^2-4ac=6^2-4*2*3=36-24=12$

As, $Delta>0$ , there are 2 real roots

$x_1=(6-sqrt12)/(2*2)=(6-2sqrt3)/(4)=(3-sqrt3)/(2)$

$x_2=(6+sqrt12)/(2*2)=(6+sqrt3)/(4)=(3+sqrt3)/(2)$

Let our inequality be

$f(x)=(x-x_1)(x-x_2)$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $x_1$ $color(white)(aaaa)$ $x_2$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-x_1$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-x_2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

$f(x)>=0$ when $x in (-oo,(3-sqrt3)/2] uu [(3+sqrt3)/2,+oo)$
graph{2x^2-6x+3 [-4.93, 4.934, -2.465, 2.465]}

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