How do you solve #16- 4x ^ { 2} = - 20#?

1 Answer
Jun 7, 2017

Answer:

#x=+-3#

Explanation:

#"subtract 16 from both sides"#

#cancel(16)cancel(-16)-4x^2=-20-16#

#rArr-4x^2=-36larr" multiply through by - 1"#

#rArr4x^2=36#

#"divide both sides by 4"#

#(cancel(4)color(white)(x)x^2)/cancel(4)=36/4#

#rArrx^2=9#

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

#sqrt(x^2)=+-sqrt9larrcolor(red)" note plus or minus"#

#rArrx=+-3#