How do you solve #-2k ^ { 2} + 8= 0#?

3 Answers
Caitlyn D. · Jumbotron
Apr 24, 2018

#k=+2 or -2#

Explanation:

#-2k^2+8=0#

#-2k^2+8-8=0-8#

#-2k^2=-8#

#((-2k^2)/-2)=-8/-2#

#k^2=4#

#sqrt(k^2)= +-sqrt(4)#

#k=+-2 #

Jim G.
Apr 24, 2018

#k=+-2#

Explanation:

#"subtract 8 from both sides of the equation"#

#rArr-2k^2=-8#

#"divide both sides by "-2#

#rArrk^2=4#

#color(blue)"take the square root of both sides"#

#rArrk=+-sqrt4larrcolor(blue)"note plus or minus"#

#rArrk=+-2#

Meave60
Apr 24, 2018

#k=-2,2#

Explanation:

Solve:

#-2k^2+8=0#

Subtract #8# from both sides.

#-2k^2=-8#

Divide both sides by #-2#.

#k^2=(-8)/(-2)#

#k^2=8/2#

#k^2=4#

Take the square root of both sides.

#sqrt(k^2)=+-sqrt4#

Apply rule #sqrt(a^2)=a#.

#k=+-2#

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