You are here:



For the life of me, I cannot figure this one out.  A step by step  would be great....

(x-1)^.5    + t = (x/2)^.5

solve for x.

I believe the next step is

sqrt(x-1)     + t  = sqrt (x/2)

But I'm stuck after this step.

This is a problem that requires a lot of attention to detail.
Once you have sqrt(x-1) + t  = sqrt (x/2), subtract sqrt(x-1) from both sides.
This gives t = sqrt(x/2) - sqrt(x-1).

Now that we have t on one side and x on the other, square both sides.
This gives t = x/2  - 2*sqrt(x/2)*sqrt(x-1) + x-1.

Combining x/2 and x-1 gives 3x/2 - 1, so this makes it t = 3x/2 - 1 - 2*sqrt(x/2)*sqrt(x-1).

This time we separate the square root to one side by adding -3x/2 + 1 to both sides,
giving t - 3x/2 + 1 = -2*sqrt(x/2)*sqrt(x-1).
This is the same as t - (3x/2 - 1) = -2*sqrt(x/2)*sqrt(x-1).

If we square this again, we get t^4 - 2t(3x/2 - 1) + (3x/2 - 1) = 4(x/2)(x-1).

Multiplying out the entire equation gives
t^4 - 3tx + 2t + 9x/4 - 3x + 1 = 2x - 2x.

Adding -2x + 2x and (noting that -2x is really -8x/4) to both sides gives
t^4 - 3tx + 2t + x/4 - x + 1 = 0.

Arranging the left side in terms of x gives
x/4 + -(3t + 1)x + t^4 + 2t + 1 = 0.

This is a quadratic equation a = 1/4, b = -(3t + 1), and c = t^4 + 2t + 3.

[As a comment, check both of the roots out since one of them is probably not an answer.
The reason for this is that we squared both sides, and that produces the extraneous root.
Trying to figure out which root this is could be a bit of a challenge.]

The quadratic equation is x = (-b sqrt[b-4ac])/(2a).
This is going to get bad since we have rather messy expressions for b and c.
Note that a = 1/4, so 2a = 1/2, and dividing by 1/2 is the same as multiplying by 2.

Using this, it gives x = 2([3t + 1] sqrt([3t + 1] -  t^4 - 2t - 3).
Note this can be reduced a little to   x =  2([3t + 1] sqrt[5t^4 + 4t - 2]).

I've been through this problem a half dozen times and found a correction every time,
but it seems like its right so far.  I won't even go into putting it back in to see
if any of the roots are extraneous.  


All Answers

Answers by Expert:

Ask Experts


Scott A Wilson


Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.


I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

Documents at Boeing in assistance on the manufacturiing floor.

MS at math OSU in mathematics at OSU, 1986. BS at OSU in mathematical sciences (math, statistics, computer science), 1984.

Awards and Honors
Both my BS and MS degrees were given with honors.

Past/Present Clients
Students in a wide variety of areas since the 80's; over 1,000 of them have been in algebra.

©2016 All rights reserved.