solving radical equations: square root of 2y + 7, + 4 = y

[jburgswife30]

the square root of 2y+7, +4=y

here is what I have but I think I messed up somewhere

2y+7=y-4

(2y+7)^2=(y-4)^2
2y+7=(y-4)(y-4)
2y+7=y^2-4y-4y+16
2y+7=y^2-8y+16
2y=y^2-8y-9
0=y^2-10y-9

 

[stapel]

the square root of 2y+7, +4=y

I'm going to guess that the comma indicates the end of what is inside the radical, so the equation is as follows:

. . . . .sqrt[2y + 7] + 4 = y

I'm not sure what happened in your work, because the square root appears to have disappeared...?

I think you were on the right track, though: For this sort of equation, a good first step is to isolate the radical (get it by itself), and then square both sides (since squaring "undoes" square-rooting):

. . . . .sqrt[2y + 7] = y - 4

. . . . .( sqrt[2y + 7] )² = (y - 4)²

. . . . .2y + 7 = y² - 8y + 16

Get everything over on one side (the right-hand side looks easier):

. . . . .0 = y² - 8y - 2y + 16 - 7

...simplify, and then solve the resulting quadratic equation.

Eliz.

 

[Deleted member 4993]

Re: solving radical equations: square root of 2y + 7, + 4 =

the square root of 2y+7, +4=y

Is the problem:

\(\displaystyle \L \sqrt{2y\,+\,7\,}\, +\, 4\, =\, y\)

 

[jburgswife30]

yes it is

 

[Deleted member 4993]

Stapel has already anticipated the correct form and corrected your solution above.