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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 t² = 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 t² = 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 t² - 3x/2 + 1 = -2*sqrt(x/2)*sqrt(x-1).

This is the same as t² - (3x/2 - 1) = -2*sqrt(x/2)*sqrt(x-1).

If we square this again, we get t^4 - 2t²(3x/2 - 1) + (3x/2 - 1)² = 4(x/2)(x-1).

Multiplying out the entire equation gives

t^4 - 3t²x + 2t² + 9x²/4 - 3x + 1 = 2x² - 2x.

Adding -2x² + 2x and (noting that -2x² is really -8x²/4) to both sides gives

t^4 - 3t²x + 2t² + x²/4 - x + 1 = 0.

Arranging the left side in terms of x gives

x²/4 + -(3t² + 1)x + t^4 + 2t² + 1 = 0.

This is a quadratic equation a = 1/4, b = -(3t² + 1), and c = t^4 + 2t² + 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[b²-4ac])/(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([3t² + 1] ± sqrt([3t² + 1]² - t^4 - 2t² - 3).

Note this can be reduced a little to x = 2([3t² + 1] ± sqrt[5t^4 + 4t² - 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.

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Comment | Thank you so much. When I got into that problem, I realized that it was over my head, even tho I fully understood decimal exponents. I wonder about any problem that is so convoluted. Why would it be in any book, even a college book? Pointless, I think, but then, I'm not an author of math books. Thank you so much for your time..... |

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