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

2 Answers
LM
Mar 13, 2018

r = 1 or r = 3

Explanation:

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

Ben
Mar 13, 2018

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

Explanation:

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)

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