Solve this equation?

Question

sqroot of (2x + 1) - square root of ( x ) = 1

Please explain how you got the answer if you have time. I'm preparing for a test. Thanks a bunch.

Answer 1

You should square both sides, but you might make an error like the others did. Be sure to multiply correctly using the FOIL (First, Outer, Inner, Last) method.

The reason is that [ sqrt(a) - sqrt(b) ]² does *not* equal a + b. It actually equals a - 2sqrt(a)sqrt(b) + b.

The correct equation following their method should be:
(2x + 1) - 2(sqrt(2x + 1)sqrt(x)) + x = 1

3x + 1 - 1 - 2sqrt(2x² + x) = 0
3x - 2sqrt(2x² + x) = 0

3x = 2sqrt(2x² + x)
(3x)² = 4*(x² + x)
9x² = 8x² + 4x
x² - 4x = 0
x(x - 4) = 0

x = 0 or x = 4

Note: if I were doing this, I would probably switch the sqrt(x) to the right side and then square:

sqrt(2x + 1) = 1 + sqrt(x)
2x + 1 = 1 + 2sqrt(x) + x
x = 2 sqrt(x)
x² = 4x
x² - 4x = 0
x(x-4) = 0

Either way you get to x = 0 or x =4.

And you should always double-check your answers:
CASE 1: x = 0
sqrt(2(0) + 1) - sqrt(0) =? 1
sqrt(1) - sqrt(0) =? 1
1 - 0 =? 1
1 = 1

CASE 2: x = 4
sqrt(2(4) + 1) - sqrt(4) =? 1
sqrt(9) - sqrt(4) =? 1
3 - 2 =? 1
1 = 1

So the correct answers are x = 0 or x = 4

Answer 2

First,
sqr (2x + 1) - sqr( x ) = 1
sqr (2x + 1) = 1+ sqr( x )

Now, do the squaring right
2x+1 = 1+2sqr(x) + x
x= 2sqr(x)
Now, square again
x^2 = 4 x
x=4, x=0

Now, double check by pluging x = 4 back in
sqr (2x + 1) - sqr( x ) = 1
sqr (8 + 1) - sqr( 4) = 1
sqr (9) - sqr( 4) = 1
3-2 =1

Now, double check by pluging x = 0 back in
sqr (0 + 1) - sqr( 0 ) = 1
1 = 1

Answer 3

To get rid of the square roots; square both sides

(2x+1) - x = 1

Combine like terms

x +1 = 1

Subtract 1 from both sides
x = 0

Check:

sqroot of (2*0+1) - square root of 0 = 1
sqroot of 1 - 0 = 0

sqroot of 1 is 1 so your x is 0

Answer 4

There is some shady algebra goin' on....

[sqrt(2x+1) - sqrt(x)] squared is

2x+1 + 2sqrt[x(2x+1)] + x

so after squaring you should get

3x + 1 + 2sqrt[x(2x+1)] = 1.

Now, isolate the radical, then square again. Then, solve for x. Don't forget to check your answers! Good luck.

Answer 5

sqrt(x) = sqrt(2x+1) - 1
x = (2x+1) - 2sqrt(2x+1) + 1
-x = 2 - 2sqrt(2x+1)
x = 2sqrt(2x+1) - 2
x = 2 * (sqrt(2x+1)-1)
remember sqrt(x) = sqrt(2x+1) -1
x = 2 * sqrt(x)
x^2 = 4x
x^2 - 4x = 0
x(x-4) = 0

x = 0, x = 4


NOTE: Answers above me are incorrect. You can't just square the left side to get rid of all the sqrt's. It turns into a FOIL.

Answer 6

sqrt(2x + 1) - sqrt(x) = 1

Firstly, square both sides so you do not have to deal with the radical.

2x + 1 - x = 1

Next, simplify.

x = 0

So x = 0.

Plug that in to make sure it's correct.

sqrt(0 + 1) - sqrt(0) = 1

Since sqrt(1) is 1, and 1 = 1 makes a true statement, it's correct.

Hope I helped!

Answer 7

Just by looking and taking a guess, I would say that x = 0 is a possible answer.

Answer 8

mathguru is correct. when squaring both sides, use FOIL METHOD. it is like (a-b) = 1

you wont say a^2 - b^2 = 1^2

it just doesnt cancel the radical signs geez

it would more likely be

a^2 - 2ab - b^2 = 1



get it?

Answer 9

If you are preparing for an exam and asking for someone on Yahoo to do your work instead of learning and practicing the problems yourself, you are in deep stuff.