sqrt(x) + sqrt(2x-1) - 2 = 0?

Question

sqrt(x) + sqrt(2x-1) - 2 = 0.

plz show me how to solve this

Answer 1

x = 1 is a clear solution

I don't think there are other solutions.

EDIT:
citsy. That's not how you square. Remember that the entire thing is being squared. Not just the individual expressions.

My method? I'm use to guess and check so I'm getting quite skilled at it. It's useful for factoring large polynomials (unless the roots are absurd lol).

1 could have been the only solution because The rest wouldn't have been integers.

When you observe squares, you'll understand.

Santmann's method I never would have thought of, because it would definitely that much thinking and time. That could have been the only mathematical way to do it though.

Answer 2

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

Subtract sqrt(2x - 1) from both sides:

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

Square both sides:

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

Subtract 4 from both sides:

x - 4 sqrt (x) = 2x - 5

Subtract x from both sides:

-4 sqrt (x) = x - 5

Square both sides again:

16x = (x - 5)(x - 5)
16x = x^2 - 10x + 25

Subtract 16x:

0 = x^2 - 26x + 25

Factor:

0 = (x - 25)(x - 1)

Potential solutions: x = 1 and x = 25.

Since squaring both sides of any equation may introduce extraneous solutions, you must *check* both of these solutions to see if they satisfy the original equation.

x = 1:
sqrt(1) + sqrt(2(1) - 1) - 2 = 0: This is true. So x = 1 is a solution.

sqrt(25) + sqrt(2(25) - 1) - 2 = 0: This is not true. So x = 25 is not a solution.

So the only solution is x = 1.

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Note: You can solve problems like this one, with two square roots in them, by the following process:

1) Get the two square roots on opposite sides
2) Square both sides
3) There will be only one square root left. Get it by itself on one side of the equation.
4) Square both sides.
5) Now solve (using methods for solving a quadratic equation).

Answer 3

move the -2 to the other side

sqrt(x) + sqrt(2x-1) = 2
square everything
x + 2x - 1 = 4
combine x's
3x - 1 = 4
move 1 to other side
3x = 5
divide by 3
x = 5/3


*Edit* I just read the first person's answer...and hes right that one does work. I have no idea how he got that because I solved the way one usually would...but one is the answer...or at least ONE of the answers (could be more). I still think my method is right, perhaps I made a simple math error somewhere.

Answer 4

If sqrtx = sqrt(2x-6), then x = 2x - 6 and x = 6 If sqrtx = (sqrt2x)-6, then x = 2x - 12(sqrt2x) + 36 and x = 12(sqrt2x) -36 = 12(sqrt2x - 3), from which i'm getting x = 6.176623. by some skill I doubt it is what they needed.

Answer 5

sqrt(x) +sqrt(2x-1)=2//x>=1/2
square both sides
x+2x-1 +2 sqrt(x)*sqrt(2x-1) =4
2 sqrt(x)*sqrt(2x-1) = 5-3x // 5-3x>=0 so x<=5/3
with those conditions, square again
4x(2x-1) = 25 -30x+9x^2
x^2-26x+25=0 so x= ((26+- sqrt(576))/2

x= 25 outside the range and x=1 only solution

Answer 6

sqrt(x) + sqrt(2x-1) - 2 = 0
sqrt(2x+1) = 2 -sqrt(x) then square
2x+1 = 4 -4sqrt(x) +x
3-x=4sqrt(x) then square
x^2-6x+9 =16x -> x^2-22x+9=0 solve quadratic equation
x= [-(-22) +-sqrt(22^2-4*1*9)]/2 = 21.58 or 0.417

good luck

Answer 7

o thats easy just find wat the number of the sqrt is then multiply it by the number of the X then add the number of sqrtmultiply that by 2 then divide that byt the number of X minus 1-2=0

Answer 8

Subtract 2 from both sides and solve for x.