For which value of k will the roots of #2x^2+kx+1=0# be real?

1 Answer
Dec 2, 2016

Answer:

The answer is #k in ] -oo,-sqrt8 ] uu [sqrt8, oo[ #

Explanation:

In order for the roots of a quadratic equation, to de real, the dircriminant #Delta>=0#

If the equation is #ax^2+bx+c=0#

the discriminant is #Delta=b^2-4ac#

#2x^2+kx+1=0#

#Delta= k^2-4*2*1>=0#

#k^2-8>=0#

#(k-sqrt8)(k+sqrt8)>=0#

We do a sign chart

#color(white)(aaaa)##k##color(white)(aaaa)##-oo##color(white)(aaaa)##-sqrt8##color(white)(aaaa)##sqrt8##color(white)(aaaa)##+oo#

#color(white)(aaaa)##k+sqrt8##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##k-sqrt8##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##Delta##color(white)(aaaaaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaa)##+#

So, for #Delta>=0#

#k in ] -oo,-sqrt8 ] uu [sqrt8, oo[ #