How do you solve for sqrt(4s+17)-s-3=0?

2 Answers
May 3, 2018

s=2

Explanation:

sqrt(4s+17)-s-3=0
sqrt(4s+17)=s+3
(sqrt(4s+17))^2=(s+2)^2
4s+17=s^2+6s+9
s^2+6s+9-4s-17=0
s^2+2s-8=0

D=2^2-4*1*(-8)=4+32=36

s_1=(-2+6)/2=2
s_2=(-2-6)/2=-4

-4 is not the solution of the equation. Because square root cannot be negative. (s+3)=(-4+3)=-1

2 is the correct answer:
sqrt(4*2+17)=2+3
sqrt25=5
5=5

May 3, 2018

s=2

Explanation:

"subtract "-s-3" from both sides of the equation"

rArrsqrt(4s+17)=s+3

color(blue)"square both sides"

(sqrt(4s+17))^2=(s+3)^2larrcolor(blue)"distribute"

rArr4s+17=s^2+6s+9

"collect like terms and equate to zero"

s^2+2s-8=0larrcolor(blue)"in standard form"

"the factors of - 8 which sum to + 2 are + 4 and - 2"

rArr(s+4)(s-2)=0

"equate each factor to zero and solve for s"

s+4=0rArrs=-4

s-2=0rArrs=2

color(blue)"As a check"

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

s=-4tosqrt(-16+17)+4-3=1+4-3=2

2!=0rArrs=-4" is extraneous"

s=2tosqrt(8+17)-2-3=sqrt25-5=5-5=0

rArrs=2" is the solution"