Solve by using the quadratic formula?

Question

#3k^2 - 18k + 22 = 0#

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Answer 1 · 2018-04-25T18:40:32

#k=3+-sqrt15/3#

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If we have a quadratic function in the form of:

#ax^2+bx+c=0#, then

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#k=(-(-18)+-sqrt((-18)^2-4(3)(22)))/(2(3))=(18+-sqrt(324-264))/6=(18+-sqrt60)/6=(18+-2sqrt15)/6=(9+-sqrt15)/3#

#k=3+-sqrt15/3#

Answer 2 · 2018-04-25T18:42:40

#k=3+-sqrt15/3#

Here,

#3k^2-18k+22=0#

Comparing with #ax^2+bx+c=0#,we get

#a=3, b=-18and c=22#

#triangle=b^2-4ac=324-4(3)(22)=324-264=60#

#sqrt(triangle)=sqrt60=2sqrt15#

So,

#k=(-b+-sqrt(triangle))/(2a)=(18+-2sqrt15)/(2xx3)=18/6+- (2sqrt15)/6#

#:.k=3+-sqrt15/3#