How do you solve for sqrt(4s+17)-s-3=0?
2 Answers
Explanation:
-4 is not the solution of the equation. Because square root cannot be negative.
2 is the correct answer:
Explanation:
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"