Question

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

Answer 1

`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`

Answer 2

`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"`