How do you solve #r= \sqrt { - 3+ 4r }#?

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Answer 1 · 2018-03-13T21:50:37

#r = 1 or r = 3#

#r = sqrt(-3+4r)#

square both sides:

#r^2 = -3 + 4r#

add #3# and subtract #4r#, to make the RHS #0:#

#r^2 - 4r + 3 = 0#

factorise the quadratic equation:

#-1 + -3 = -4#
#-1 * -3 = 3#

#r^2-4r+3 = (r-1)(r-3)#

#(r-1)(r-3) = 0#

#r-1=0 or r-3 = 0#

#r = 1 or r = 3#

Answer 2 · 2018-03-13T21:52:31

Square both sides, then use the quadratic formula to find #r=1# and #r=3#

First, square both sides to remove the radical:

#r^2=(sqrt(-3+4r))^2 rArr r^2=-3+4r#

Next, move all terms to the left hand side to get a quadratic equation:

#r^2-4r+3=0#

fill in the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Where #a=1#, #b=-4#, and #c=3#

#x=(4+-sqrt(16-12))/2 rArr x=(4+-2)/2#

This give you two solutions, #color(red)(x=3)# and #color(blue)(x=1)#