How do you solve #2x^2+4x<=3# using a sign chart?

1 Answer
Jan 4, 2017

Answer:

The answer is #x in [-2.58, 0.58]#

Explanation:

Let`s rewrite the equation

#2x^2+4x-3<=0#

We need the roots of the equation

#2x^2+4x-3=0#

We calculate the discriminant

#Delta=b^2-4ac=16+4*2*3=40#

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

Therefore,

#x_1=(-4+sqrt40)/4=0.58#

#x_2=(-4-sqrt40)/4=-2.58#

Let #f(x)=(x+2.58)(x-0.58)<=0#

Now, we can do our sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2.58##color(white)(aaaaa)##0.58##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2.58##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##x-0.58##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaaaa)##+#

Therefore,

#f(x)<=0#, when #x in [-2.58, 0.58]#