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

2 Answers
May 3, 2018

Answer:

#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

Answer:

#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"#