sqrt 2y+7 +4=y
2007-12-17 15:02:45 · 4 answers · asked by God Bless America!~ 4 in Science & Mathematics ➔ Mathematics
sqrt of 2y+7 +4=y
sqrt of 2y+7=y-4
square the whole problem
2y+7=y^2-8y+16
y^2-10y+9=0
(y-1)(y-9)=0
y=1 and 9
2007-12-17 15:17:47 · answer #1 · answered by Dave aka Spider Monkey 7 · 2⤊ 0⤋
â (2y + 7) + 4 = y
=> â (2y + 7) = y -- 4
=> (2y + 7) = (y -- 4)^2
=> 2y + 7 = y^2 -- 8y + 16
=> y^2 -- 10y + 9 = 0
=> (y -- 1)(y -- 9) = 0
giving y = 1, 9
2007-12-17 23:12:12 · answer #2 · answered by sv 7 · 2⤊ 0⤋
Square both sides: I'm going to assume 4 is not under the radical, as you left them seperate.
2y+7=(y-4)^2
2y+7=y^2-8y+16
y^2-10y+9=0
(y-9)(y-1)
y=9, 1
2007-12-17 23:07:54 · answer #3 · answered by Bushman 2 · 2⤊ 0⤋
Hi. Get the square root on one side of the equation and everything else on the other. Square both sides and then send all terms to one side. Solve that quadratic equation. Check both answers in the original equation to make sure that they work, as any time you square both sides of an equation, it is possible to create an extra root that doesn't work in the original. Good luck!
2007-12-17 23:07:21 · answer #4 · answered by not sixty yet 2 · 0⤊ 2⤋