Question

sqrt(4y+5) - sqrt(y-1) = 3
y=??

Answer 1

not it...
I think we are looking for two different values for the y of each side? Perhaps I'm wrong...

Answer 2

how'd you get 3+(y-1)

if you squared the one side it'd be

(3+sqroot(y-1))^2

Answer 3

koko has the right idea
just be careful squaring the right side

Answer 4

you should get
3^2 + 2(3sqrt(y-1)) + sqrt(y-1)^2
= 9+6sqrt(y-1) + (y-1)
=8 + 6sqrt(y-1) + y

Answer 5

\[\sqrt{4y+5}=3+\sqrt{y-1}\] square both sides
\[4y+5=9+6\sqrt{y-1}+(y-1)\] simplify
\[3y-3=6\sqrt{y-1}\] square again
\[9y ^{2}-18y+9=36(y-1)\] simplify
\[9y ^{2}+18y+45=0\]
then factor or use the quadradic

Answer 6

\[\large \sqrt{4y+5}-\sqrt{y-1}=3\]


\[\large \sqrt{4y+5}=3+\sqrt{y-1}\]



\[\large 4y+5=(3+\sqrt{y-1})^2\]



\[\large 4y+5=9+6\sqrt{y-1}+y-1\]



\[\large 4y+5=8+y+6\sqrt{y-1}\]



\[\large 4y+5-8-y=6\sqrt{y-1}\]


\[\large 3y-3=6\sqrt{y-1}\]



\[\large (3y-3)^2=36(y-1)\]



\[\large 9y^2-18y+9=36(y-1)\]



\[\large 9y^2-18y+9=36y-36\]



\[\large 9y^2-18y+9-36y+36=0\]



\[\large 9y^2-54y+45=0\]



\[\large 9(y^2-6y+5)=0\]



\[\large 9(y-1)(y-5)=0\]



\[\large y-1=0 \ \ \textrm{or} \ \ y-5=0\]



\[\large y=1 \ \ \textrm{or} \ \ y=5\]

Answer 7

WOW! I'm in over my head!!
thnk you:)