What are the set of values for which this equation has real distinct roots?

#2x^2 + 3kx +k=0#

2 Answers
Apr 18, 2018

#k<0" or "k>8/9#

Explanation:

#"to determine the nature of the roots use the "color(blue)"discriminant"#

#•color(white)(x)Delta=b^2-4ac#

#• " If "Delta>0" then real distinct roots"#

#• " If "Delta=0" then real and equal roots"#

#• " If "Delta<0" then complex roots"#

#"here "Delta>0" is required"#

#2x^2+3kx+k=0larrcolor(blue)"is in standard form"#

#"with "a=2,b=3k" and "c=k#

#rarrDelta=(3k)^2-(4xx2xxk)=9k^2-8k#

#"rArr9k^2-8k>0#

#"the left side is a quadratic with positive leading"#
#"coefficient and zeros at "k=0" and "k=8/9#
graph{9x^2-8x [-10, 10, -5, 5]}

#"Thus it is positive when "k<0" or "k>8/9#

#k in(-oo,0)uu(8/9,oo)#

Apr 18, 2018

#k<0##" , "##k>8/9#

Explanation:

#2x^2+3kx+k=0#
is a quadratic equation

and to find the roots of #x# of quadratic equations

#color(green)"Example :"##color(blue)(ax^2+bx+c=0)#

we use the following formula

#color(blue)(x=(-b+-sqrt(b^2-4ac))/(2a)#
#color(blue)("where the term "(b^2-4ac) " is the discriminant"#

so in the quadratic equation

If the discriminant
#b^2-4ac>0##rarr#the equation has two real solutions .
#b^2-4ac<0##rarr#the equation has no real solutions .
#b^2-4ac=0##rarr#the equation has one real solution .

#2x^2+3kx+k=0#

#a=2##" , "##b=+3k##" , "##c=k#

Substitute in the discriminant

#9k^2-(4)(2)(k)#

so in order to get the real distinct roots of the function

#color(blue)("discriminant">0#

#9k^2-8k>0#

#k(9k-8)>0#

#k<0##" , "##k>8/9#