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"#