Honors pre-calculus homework, help

[hpdwnsn95]

honors pre-calculus homework, help!

Homework Statement

find the sum of the roots of the equation (x-1)^1/2 + (2x-1)^1/2 = x

Homework Equations

I have no idea, I just started a pre calculus course about 5 weeks ago and our teacher gave the people in honors problems we've never seen before so I'm not sure what would be a relevant equation

The Attempt at a Solution

I know I need to put the problem in standard form before i can find the roots, but would that look like this:(x-1)^1/2 + (2x-1)^1/2 -x = y ? I'm not sure exactly how to start this one...

 

[8point1]

How might you get rid of the square roots?

 

[symbolipoint]

1. Homework Statement
find the sum of the roots of the equation (x-1)^1/2 + (2x-1)^1/2 = x

To find the roots, you first should use algebraic properties to transform the given equation into a polynomial equation of degree 2 (meaning quadratic equation; something times something squared plus something else times the something plus a constant equals zero). Knowing how would be a result of your intermediate algebra. Once you find the roots, the solutions to the original equation, just finish by adding them, according to the given instructions, "find the sum of the roots.."

 

[hpdwnsn95]

ok thanks both of you!

 

[NascentOxygen]

What steps did you go through to solve this, hpdwnsn95?

 

[symbolipoint]

ok thanks both of you!

What steps did you go through to solve this, hpdwnsn95?

If he said he solved it, then I believe it. I also solved it myself. Answer was "2". Not difficult. One false root occurred in the solution process, eliminated by division of both members of the equation. Two "roots" remaining, easily found and processed.

 

[LCKurtz]

If he said he solved it, then I believe it. I also solved it myself. Answer was "2". Not difficult. One false root occurred in the solution process, eliminated by division of both members of the equation. Two "roots" remaining, easily found and processed.

Interesting that you got 2 for the sum of the roots when they are x = 1 and x = 5.

 

[NascentOxygen]

If he said he solved it, then I believe it. I also solved it myself. Answer was "2". Not difficult.

I had difficulty understanding the suggested method/s of solving this, and hoped hpdwnsn95's answer would clear up what I was missing. But thank you for butting in just the same.

 

[LCKurtz]

I had difficulty understanding the suggested method/s of solving this, and hoped hpdwnsn95's answer would clear up what I was missing. But thank you for butting in just the same.

Me too. I was thinking the problem might be suggesting some clever way to find the sum of the roots without brute force squaring it out. Once you have the roots it seems pretty silly to ask what the sum of them is.

 

[symbolipoint]

Interesting that you got 2 for the sum of the roots when they are x = 1 and x = 5.

With that analysis, my own work deserves more careful checking. My fourth attempt seems to show this set of steps to reach the quadratic equation stage of the solution:

\[<br /> \begin{array}{l}<br /> (x - 1)^{1/2} + (2x - 1)^{1/2} = x \\ <br /> (x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} = x - (2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} \\ <br /> x - 1 = [x - (2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} ]^2 \\ <br /> x - 1 = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} + 2x - 1 \\ <br /> x - 1 - 2x + 1 = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} \\ <br /> - x = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} \\ <br /> - x - x^2 = - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} \\ <br /> x + 1 = 2(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}} \\ <br /> x^2 + 2x + 1 = 4(2x - 1) \\ <br /> x^2 + 2x + 1 = 8x - 4 \\ <br /> x^2 - 6x + 5 = 0 \\ <br /> \end{array}<br /> \]<br />

Remaining to be done is find the actual roots and then their sum.

 

[NascentOxygen]

Me too. I was thinking the problem might be suggesting some clever way to find the sum of the roots without brute force squaring it out.

Squaring it out twice, at that.

But once you have x² - 6x + 5
remember...

-(sum of roots) = -6, and
product of roots = +5

Though, given this is a pre-calculus exercise, I think the recommended first option should have been to try the whole numbers from, say, -10 to +10. We can scratch all the negatives, since RHS must be positive as it results from addition, then one at a time try from 1..upwards.

 

[LCKurtz]

Squaring it out twice, at that.

But once you have x² - 6x + 5
remember...

-(sum of roots) = -6, and
product of roots = +5

Yes, I knew that. But if you have to go through the work required to get it to there, finding the roots is so trivial that being asked to find the sum of the roots is pretty much pointless. Now if there were some clever way of getting the sum of the roots from the original equation, without finding the roots, that would would make it an interesting problem. And since you are asked for the sum of the roots and not the roots themselves, I wondered if there wasn't some clever trick that I haven't seen before.