Question

√y+5 + √y +13 = 4 (the radical is over y+5 and y+13)

Can someone help me with this

Answer 1

The first thing you should try to do is eliminate the radicals over the y's. There's usually two ways you can do this: multiplying it with the conjugate or squaring both sides.

Answer 2

the radicals are over both y+5 and y +13 not only y

Answer 3

Yes, I understand. You can still do it this way, though.

Answer 4

You cant just square root y and leave out the 5 or the 13

Answer 5

The most sensible thing to do is to square both sides like so:
\[(\sqrt{y+5} + \sqrt{y+13})^2 = 4^2\]

Answer 6

This is what I mean. You square both sides.

Answer 7

Ok I see where you're going

Answer 8

You'll get something like:
\[18+2y+2(\sqrt{y+5}+\sqrt{y+13})=16\]

Answer 9

It should be easy from this point onward.

Answer 10

No im soory i dont get it youkind of just added more stuff right?

Answer 11

What i did i subtracted the rad y +5 and then squared both sides to get rid of the radicals

Answer 12

\[(\sqrt{y+5} +\sqrt{y+13})(\sqrt{y+5} +\sqrt{y+13})=16\]

Yes?
Multiply the left hand side using the "FOIL" method.

Answer 13

Not following you....

Answer 14

Okay. Let's take it slow.
Your equation is
\[\sqrt{y+5}+\sqrt{y+13}=4\]
You have to square both sides.
\[(\sqrt{y+5}+\sqrt{y+13})^2 = 4^2\]

Answer 15

Do you understand this so far?

Answer 16

Ok but cant you also subtract the y+5 and then square it wouldnt that be easier

Answer 17

I see what you mean but isnt this faster

Answer 18

If you subtract the y+5, you would have to subtract it from BOTH SIDES. Which means you would end up with the same equation anyway.

Answer 19

The rad y+5 so it can cancel one side

Answer 20

\[\sqrt{y+5}+\sqrt{y+13}-\sqrt{y+15} = 4-\sqrt{y+5}\]

Answer 21

Is this what you mean?

Answer 22

the rad y +5 is suppose to cancel on the left side you didnt subtract it

Answer 23

Or sorry i meant the y +3

Answer 24

not y+5 i dont know where i got that one from

Answer 25

It can be solved by using the following logic:
\[0\le \sqrt{Y+5}<4\]
\[1\le \sqrt{Y+13}\le4\]
Note that Y can be a negative integer.

Answer 26

I ended up with 24 - 8√y+13 sorry kropot i dont want to use inequalities

Answer 27

is that right so far

Answer 28

I'm not really sure what you're trying to do... But this should be the solution:
\[(\sqrt{y+5}+\sqrt{y+13})^2=4^2\]
\[18+2y+2\sqrt{(y+5)(y+13)}\]
\[2\sqrt{(y+5)(y+13)}=-2-2y\]
\[4(y+5)(y+13) = (-2-2y)^2\]
\[ 4y^+72y+260=(-2-2y)^2\]
\[4y^2+72y+260=4y^2+8y+4\]
64y+256=0
64(y+4)=0
y=-4

Answer 29

Sorry, the second step should be =16

Answer 30

The fourth one should be
4y^2 + 72y

Answer 31

Let me do it on my own one more time and see what i come up with

Answer 32

Okay.

Answer 33

I got y to equal 525

Answer 34

If you plug that value into the left side of your equation, you wouldn't get the right hand side (which is 4).

Answer 35

Let me show you my work

Answer 36

and sorry this taking so long

Answer 37

\[√y+15 + √y+13 = 4
(√y+5)^{2} = (4- √y+13)^{2}
y+5 = 16 - 8√y +13 + y + 13
y+5 +29 - 8√y +13
y = 24 - 8√y +13
y = 576 -64 + y + 13\]
y = 525

Answer 38

I see the problemthe y's will cancel out and leave an empty solution

Answer 39

This solution isn't valid either. Remember that if you do something on one side, you have to do it in the other side as well. So if you multiply one side with whatever value, you have to do it on the other side.

Answer 40

The solution you've given doesn't really make any sense. At least, to me.

Answer 41

Yeah i see that now

Answer 42

Are you going to be okay now? :)

Answer 43

Yeah ill just ask my teacher tomorrow thank you anyway