How do you solve $2x^2-8x+3<0$ using a sign chart?

Question

1593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-24T19:39:54

The solution is $x in ](4-sqrt10)/2, (4+sqrt10)/2[$

We need the roots of the equation before building the sign chart

$2x^2-8x+3=0$

We calculate the discriminant

$Delta=b^2-4ac=(-8)^2-4(2)(3)$

$=64-24=40$

$Delta>0$ , there are 2 real roots

$x_1=(8-sqrt40)/4=(4-sqrt10)/2$

$x_2=(8+sqrt40)/4=(4+sqrt10)/2$

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

We can, now, 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 $x in ]x_1,x_2[$ , $=>$

$x in ](4-sqrt10)/2, (4+sqrt10)/2[$

How do you solve 2x^2-8x+3<02x^2-8x+3<0 using a sign chart?