Honors pre-calculus homework, help

In summary, the conversation was about a pre-calculus homework problem where the goal was to find the sum of the roots of the equation (x-1)^1/2 + (2x-1)^1/2 = x. The suggested method involved transforming the equation into a quadratic equation and then finding the roots. However, one person questioned the point of finding the sum of the roots when the roots themselves were easy to find. Another person suggested a possible clever way to find the sum of the roots using algebraic properties.
  • #1
6
0
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...
 
Physics news on Phys.org
  • #2


How might you get rid of the square roots?
 
  • #3


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.."
 
  • #4


ok thanks both of you!
 
  • #5


What steps did you go through to solve this, hpdwnsn95?
 
  • #6


hpdwnsn95 said:
ok thanks both of you!

NascentOxygen said:
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.
 
  • #7


symbolipoint said:
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.
 
  • #8


symbolipoint said:
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.
 
  • #9


NascentOxygen said:
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.
 
  • #10


LCKurtz said:
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:

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

Remaining to be done is find the actual roots and then their sum.
 
  • #11


LCKurtz said:
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.
 
  • #12


NascentOxygen said:
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.
 

1. What is the purpose of honors pre-calculus?

Honors pre-calculus is an advanced high school mathematics course that covers topics in algebra, trigonometry, and pre-calculus. Its purpose is to prepare students for more advanced math courses in college, such as calculus.

2. How is honors pre-calculus different from regular pre-calculus?

Honors pre-calculus covers the same topics as regular pre-calculus, but in more depth and with a faster pace. It also includes additional challenging topics and problems to push students to their full potential.

3. What are some key skills needed for success in honors pre-calculus?

Students in honors pre-calculus should have a strong foundation in algebra, geometry, and trigonometry. They should also have good critical thinking and problem-solving skills, as well as the ability to work independently.

4. How can I get help with my honors pre-calculus homework?

There are many resources available for help with honors pre-calculus homework. You can ask your teacher for clarification and guidance, join a study group with classmates, or seek help from online tutoring services or peer tutoring programs.

5. How can I prepare for honors pre-calculus exams?

To prepare for honors pre-calculus exams, it is important to attend all classes, take thorough notes, and complete all homework and practice problems. You can also review old quizzes and tests, work on additional practice problems, and seek help from your teacher or peers if needed.

Suggested for: Honors pre-calculus homework, help

Replies
9
Views
1K
Replies
11
Views
1K
Replies
6
Views
1K
Replies
16
Views
2K
Replies
14
Views
1K
Replies
17
Views
2K
Replies
8
Views
1K
Back
Top