Solving Radical equations: sqrt(y + 1) + sqrt(y - 4) = 5

[cjcapta]

I need help on this problem to see if I am doing something wrong or if this question is one with no solution. The directions say: Solve each equation.
Original problem: ?(y+1) + ?(y-4) = 5
Step 1 square both sides. (y+1) + (y-4) = 25
Step 2 combine numbers. 2y - 3 = 25
Step 3 add 3 to both sides. 2y = 28
Step 4 divide both sides by 2. y=14

When I go to check it, I get ?(15) + ?(10) = 5
However this is not true.

P.S. On a side note, how do you make an equal sign with a line through it meaning does not equal?

 

[Deleted member 4993]

Re: Radical equations

I need help on this problem to see if I am doing something wrong or if this question is one with no solution. The directions say: Solve each equation.
Original problem: ?(y+1) + ?(y-4) = 5
Step 1 square both sides. (y+1) + (y-4) = 25<<<<< This is not correct. The step of squaring is correct - but the result is not.

remember

(a+b)[sup:1u4y53zk]2[/sup:1u4y53zk] = a[sup:1u4y53zk]2[/sup:1u4y53zk] + b[sup:1u4y53zk]2[/sup:1u4y53zk] + 2ab ...... you are forgetting the '2ab' part

 

[cjcapta]

Re: Radical equations

But it's the square root and any time you square a square root of a number you get the number inside the radical.

So ?(2) ^2 = 2 and ?(4)^2 = 4 and ?(0)^2 = 0 and ?(x+3)^2 = x+3 and ?(3+x+2x) = 3+x+2x or then 3+3x.
(I choose these numbers from random.)

 

[mmm4444bot]

Re: Radical equations

But it's the square root ...

No, CJ, it is not.

It is the SUM of TWO square roots which you are squaring.



If you were simply squaring ?(y + 1), then the result would be y + 1.

If you were simply squaring ?(y - 4), then the result would be y - 4.

You ARE NOT simply squaring two square roots, individually.

You are squaring their SUM.

In other words, (A + B)^2 does not equal A^2 + B^2.

Use FOIL to multiply [?(y + 1) + ?(y - 4)] TIMES [?(y + 1) + ?(y - 4)], and this should become clear.

If you still don't see why, then please show us your work in using FOIL to square the sum of these two square roots.

Cheers,

~ Mark

 

[cjcapta]

Re: Radical equations

It is the SUM of TWO square roots which you are squaring.

If you were simply squaring ?(y + 1), then the result would be y + 1.

If you were simply squaring ?(y - 4), then the result would be y - 4.

You ARE NOT simply squaring two square roots, individually.

You are squaring their SUM.

In other words, (A + B)^2 does not equal A^2 + B^2.

Use FOIL to multiply ?(y + 1) TIMES ?(y - 4), and this should become clear.

Oh, I see now, that would explain why it worked for the others. I didn't understand because in your first post you gave only one parenthesis without the square root sign.

 

[cjcapta]

Re: Radical equations

Okay so I got 2y - 3 + 2?(y^2 -3y-4) = 25 Do I square both sides again?

 

[Deleted member 4993]

Re: Radical equations

Okay so I got 2y - 3 + 2?(y^2 -3y-4) = 25 Do I square both sides again?

Isolate the radical sign on one-side of the "equal" sign - and square it again.

 

[cjcapta]

Re: Radical equations

Okay so I got 2y - 3 + 2?(y^2 -3y-4) = 25 Do I square both sides again?

Isolate the radical sign on one-side of the "equal" sign - and square it again.

Okay I have 2?(y^2-3y-4) = 28 - 2y. Should I divide both sides by 2 before I square them? Also my mind just forgot this one part of info, if I squared 2y, would it be 4y^2 or 2y^2 even though I know that in this problem it has to be (28 - 2y)^2, just wondering.

 

[mmm4444bot]

Re: Radical equations

... I have 2?(y^2-3y-4) = 28 - 2y. Should I divide both sides by 2 before I square them? ...

I would because it's an easy simplification to do at this point, since I recognize a factor of 2 in each term on the right-hand side. But squaring first and dividing by 4 later will lead to the same end result.

... if I squared 2y, would it be 4y^2 or 2y^2 ...

It will be 4y^2 because squaring a product of factors results in each factor being squared.

[2y]^2 = 2^2 * y^2

[(34xy)(1/z)]^2 means each factor gets squared:

(34^2) * x^2 * y^2 * (1/z)^2

Cheers,

~ Mark

 

[cjcapta]

Re: Radical equations

I would because it's an easy simplification to do at this point, since I recognize a factor of 2 in each term on the right-hand side. But squaring first and dividing by 4 later will lead to the same end result.

Cheers,

~ Mark [/color]

Okay, so now I have ?(y^2 -3y-4) = 14-y. If I am correct I square both sides, but does 14 - y become 14^2-y^2 or (14-y) (14-y)? I'm thinking the latter but just want to be sure. If I do, then I have y^2 - 3y -4 = 196 - 28y + y^2. If I subtract y^2 from each side, then I get -3y -4 = 196 -28y. Then I get 25y = 200. y = 8. Is this correct?

 

[Deleted member 4993]

Re: Radical equations

y = 8. Is this correct?

Put this value back in your original equation and see if you get back the identity.

 

[cjcapta]

Re: Radical equations

y = 8. Is this correct?

Put this value back in your original equation and see if you get back the identity.

I checked it, but I was just wondering if my method might have been wrong or not. Sometimes, I get a problem right, but used the wrong method, and it just turned out to be right by coincidence.

 

[mmm4444bot]

Re: Radical equations

... but does 14 - y become 14^2-y^2 or (14-y) (14-y)? I'm thinking the latter ...

(A + B)^2 does not equal A^2 + B^2

For the same reason, the following is also true.

(A - B)^2 does not equal A^2 - B^2

-

... I didn't understand because in your first post you gave only one parenthesis without the square root sign.

Huh?

-

... but I was just wondering if my method might have been wrong or not. Sometimes, I get a problem right, but used the wrong method, and it just turned out to be right by coincidence.

Your work is spread across multiple posts, embedded in questions and comments, and I'm too lazy to piece it together. If you would like me to check over your steps, then retype your work, and I will be happy to examine it.

Cheers,

~ Mark