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

1 Answer
Feb 24, 2017

Answer:

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[#