How do you solve the equation #2(x^2-5)=-x^2-1#?

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Answer 1 · 2017-06-08T21:14:25

#x=+-sqrt(3)#

Multiply the #2# through on the left hand side:

#2(x^2-5)=-x^2-1#

#2x^2-10=-x^2-1#

Add #color(red)(x^2)# and #color(blue)(10)# to both sides

#2x^2color(red)(+x^2)-10color(blue)(+10)=-x^2color(red)(+x^2)-1color(blue)(+10)#

#3x^2=9#

Divide both sides by #3#

#(3x^2)/color(red)(3)=9/color(red)(3)#

#x^2=3#

Take #sqrt(color(white)(aa))# of both sides

#color(red)(sqrt(color(black)(x^2))=color(red)(+-sqrt(color(black)(3))#

#x=+-sqrt(3)#