+0

# how would i solve this radical equation?

0
107
1

$${ \sqrt{5r-9} -3 = \sqrt{r + 4} - 2}$$

Jul 20, 2018

#1
0

Solve for r:
sqrt(5 r - 9) - 3 = sqrt(r + 4) - 2

sqrt(5 r - 9) = sqrt(r + 4) + 1

Raise both sides to the power of two:
5 r - 9 = (sqrt(r + 4) + 1)^2

(sqrt(r + 4) + 1)^2 = 5 + r + 2 sqrt(r + 4):
5 r - 9 = 5 + r + 2 sqrt(r + 4)

Subtract -9 + 5 r + 2 sqrt(r + 4) from both sides:
-2 sqrt(r + 4) = 14 - 4 r

Raise both sides to the power of two:
4 (r + 4) = (14 - 4 r)^2

Expand out terms of the left hand side:
4 r + 16 = (14 - 4 r)^2
Expand out terms of the right hand side:
4 r + 16 = 16 r^2 - 112 r + 196

Subtract 16 r^2 - 112 r + 196 from both sides:
-16 r^2 + 116 r - 180 = 0

The left hand side factors into a product with three terms:
-4 (r - 5) (4 r - 9) = 0

Divide both sides by -4:
(r - 5) (4 r - 9) = 0

Split into two equations:
r - 5 = 0 or 4 r - 9 = 0

r = 5 or 4 r - 9 = 0
r = 5 or 4 r = 9

Divide both sides by 4:
r = 5 or r = 9/4

sqrt(5 r - 9) - 3 ⇒ sqrt(5×9/4 - 9) - 3 = -3/2
sqrt(r + 4) - 2 ⇒ sqrt(9/4 + 4) - 2 = 1/2:
So this solution is incorrect

sqrt(5 r - 9) - 3 ⇒ sqrt(5×5 - 9) - 3 = 1
sqrt(r + 4) - 2 ⇒ sqrt(4 + 5) - 2 = 1:
So this solution is correct

The solution is:
r = 5

Jul 20, 2018