How do you solve the quadratic #3r^2-5=-10# using any method?

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Answer 1 · 2016-12-02T04:02:08

#r=+-sqrt15/3i#

#3r^2-5=-10#
#color(white)(a^2a)+5color(white)(aa^2a)+5color(white)(aaa)#Add 5 to both sides

#3r^2=-5#

#(3r^2)/3=(-5)/3color(white)(aaa)#Divide both sides by 3

#r^2=-5/3#

#sqrt(r^2)=sqrt(-5/3)color(white)(aaa)#Square root both sides

#r=+-isqrt(5/3)color(white)(aa)#The negative inside the square root comes out as #i#.

#r=+-isqrt(5/3)*sqrt(3/3)color(white)(aaa)#Rationalize the denominator

#r=+-isqrt(15)/3=+-sqrt15/3i#