Question

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

Answer 1

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

Explanation:

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[`