How do you solve #16- 4x ^ { 2} = - 20#?
1 Answer
Jun 7, 2017
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#
