Question

Solve the question in the comments because that's where I can type equations. How to?

Answer 1

Please type in the problem you wish to solve.

Answer 2

\[\sqrt{4y+1}-\sqrt{y-2}=3\]

Answer 3

You'll need to solve for y, but before you attempt that you'll need to square both sides of this equation. First, re-write the equation as Sqrt(46+1)=3+Sqrt(y-2).

Square both sides. Unfortunately, you will still have 1 instance of Sqrt(y-2) in your result.

Isolate this Sqrt(y-2), and then square both sides again. Show all work, please.

Answer 4

yup youre right

Answer 5

Yes: I'm with Lord_Box. Again, please show your work..

Answer 6

Sqrt(4y+1)=??

Answer 7

Sorry I had to move along in the assignment and come back to this. Now to work on it.

Answer 8

Sqrt(4y+1)=3+Sqrt(y-2)

Answer 9

4y+1=9+6Sqrt(y-2)+y-2 ?

Answer 10

Good, now isolate sqrt(y-2)

Answer 11

Ok.
4y+1=7+6Sqrt(y-2)+y
4y-6=6Sqrt(y-2)+y
3y-6=6Sqrt(y-2) and I'm stuck and not even sure if I'm in right direction.

Answer 12

you're correct, now divide 6 on both sides

and then you square both sides :)

Answer 13

1/2y-1=Sqrt(y-2)???

Answer 14

yes, now square both sides (:

Answer 15

1/4y+1=y-2 ? This was the route I was on when I failed this question but it didn't quite work out.

Answer 16

(1/2y-1)^2

using (a - b)^2 = a^2 - 2ab + b^2

you made an error when doing it :)

Answer 17

\[\sqrt{4y+1}-\sqrt{y-2}=3\]
\[\sqrt{4y+1}=3+\sqrt{y-2}\]
\[(\sqrt{4y+1})^2=(3+\sqrt{y-2})^2\]
\[4y+1=(3+\sqrt{y-2})^2\]
\[4y+1=y+7+6\sqrt{y-2}\]
\[3y=6+6\sqrt{y-2}\]
\[y=\frac{ 6+6\sqrt{y-2} }{ 3 }\]
\[y=2 + 2\sqrt{y-2}\]
\[(y-2)^2=(2\sqrt{y-2})^2\]
\[y^2-4y+4=4y-8\]
\[y^2-8y+12=0\]
\[(y-6)(y-2)=0\]
\[y= 2,6\]

Im pretty sure my method isn't the most quickest, but it gets the job done. Hopefully you understand the steps I took to find the answer

Answer 18

So that means 1/4y^2-y+1? It's just confusing to me. Also thank you GoldPhenoix, that's an intersting method. I'll keep it in mind.

Answer 19

1/4y^2-y+1= y -2

1/4y^2 -2y + 3 = 0

multiply 4 on both sides to get a better coefficient

y^2 - 8y + 12 = 0

and we have a quadratic equation to solve :)

Answer 20

Thank you so much all of you!!!! I finally understand! Thank you!!!

Answer 21

Anytime! :)

Answer 22

Now also Sqrt(4y+5)-Sqrt(y-1)=3

Answer 23

Sqrt(4y+5)=3+Sqrt(y-1)
4y+5=9+6Sqrt(y-1)+y-1

Answer 24

yes, now isolate 6sqrt(y-1)

and then divide both sides by 6

and then square both sides :)

Answer 25

4y+5=8+6Sqrt(y-1)+y
3y+5=8+6Sqrt(y-1)
3y-3=6Sqrt(y-1)

Answer 26

1/2y-1/2=Sqrt(y-1)
1/4y^2-1/2y-1/4=y-1

Answer 27

y^2-2y-1=y-1
y^2-3y=0
y(y-3)=0
y={0,3} is what I got, but {1,5} is the answer. I don't understand.