Solving Quadratic Equation: 1{(1-x)^1/2+1} +1{(1+x)^1/2-1}=1/x

  • Thread starter 1/2"
  • Start date
  • Tags
    Quadratic
In summary, a quadratic equation is a mathematical expression written in the form ax^2 + bx + c = 0, containing a variable raised to the second power. To solve a quadratic equation, one can use the quadratic formula, factor the equation, or use the square root property. The roots of a quadratic equation are the values of x that make the equation true, also known as solutions or zeros. A quadratic equation can only have two roots due to the Fundamental Theorem of Algebra. Solving quadratic equations is important in various fields, allowing for the solution of real-world problems and predictions based on mathematical models.
  • #1
1/2"
99
0

Homework Statement


1 {(1-x)^1/2+1} +1{(1+x)^1/2-1}=1/x

The Attempt at a Solution


I can't got any clue how to solve it!
I tried to multiply the terms with the LCM of the denominator but bit doesn't help.
Please give me a clue!:frown:
Thank you.
 
Last edited:
Physics news on Phys.org
  • #2
Is this your equation? You really need to put parentheses around your exponents.
[tex]\sqrt{1 - x} + 1 + \sqrt{1 + x} - 1~=~1/x [/tex]
 
  • #3
Mark44 said:
Is this your equation? You really need to put parentheses around your exponents.
[tex]\sqrt{1 - x} + 1 + \sqrt{1 + x} - 1~=~1/x [/tex]

If the equation above is correct, try squaring both sides of the equation and simplifying. Then, you should be able to have nothing but square roots on the LHS. Then, square both sides again. This will allow you to get rid of the square roots. At this point, foil both sides. If you do it right, you will be left with a quartic expression that looks something like this: [tex]0=ax^{4}-bx^{2}+1[/tex]. Then, all you have to do is find the roots.
 
  • #4
The equation should be
1 /{(1-x)^1/2+1} +1/{(1+x)^1/2-1}=1/x
I am very sorry for the mistake!
 
  • #5
For starters, multiply each term on the left side by 1. For the first term, multiply by sqrt(1 - x) - 1 over itself. For the second term, multiply by sqrt(1 - x) + 1 over itself.
Next, multiply both sides by x, which will give you an equation with two radicals on one side and an integer on the other side. You should be able to solve the equation pretty quickly after that.
 
  • #6
Pretty sure this is the eq:http://img32.imageshack.us/img32/8351/mathimage.gif

Tried solving and it has no solution.
 
Last edited by a moderator:
  • #7
Nowayz said:
Pretty sure this is the eq:http://img32.imageshack.us/img32/8351/mathimage.gif

Tried solving and it has no solution.

I tried as well and couldn't find anything useful to simplify the expression.

Mark44 said:
For starters, multiply each term on the left side by 1. For the first term, multiply by sqrt(1 - x) - 1 over itself. For the second term, multiply by sqrt(1 - x) + 1 over itself.
Next, multiply both sides by x, which will give you an equation with two radicals on one side and an integer on the other side. You should be able to solve the equation pretty quickly after that.

Following Mark44's hints, I ended up with

[tex]\frac{x\left(1-x\right)^{1/2}+x}{2-x+2\left(1-x\right)^{1/2}}+\frac{x\left(1+x\right)^{1/2}-x}{2+x-2\left(1+x\right)^{1/2}}=1[/tex]

I looked at this relation graphically and found the solution (if we are only considering real numbers), but I can't figure out how to solve the equation from the point above.

To help me try an figure out if there could be a real solution, I took the equation above and replaced the 1 on the RHS with f(x). Then I asked for what value of x would f(x)=1. The important thing to realize is that if we are looking for a real solution -1≤x≤1 because of the opposing signs in the radicals. Any value of x outside of this domain will produce complex numbers, so they couldn't satisfy f(x)=1.

I'm curious to know what kind of class this problem is for. If it is a high school algebra or intro college algebra, perhaps the OP mistakenly wrote the wrong signs in the radicals. If the signs in the radicals are supposed to be the same, then the solution is +3/4 with the sign depending on the sign in the radical. Or maybe I just don't understand how to manipulate radicals well enough for this problem.

Any ideas Mark44?
 
Last edited by a moderator:
  • #8
[tex]\frac{1}{\sqrt{1 - x} + 1} + \frac{1}{\sqrt{1 + x} - 1}~=~\frac{1}{x}[/tex]

If you do exactly as I said in post 5, this equation becomes
[tex]\frac{-\sqrt{1 - x} + 1}{x} + \frac{\sqrt{1 + x} - 1}{x}~=~\frac{1}{x}[/tex]

Now multiply both sides by x. This won't cause us to lose a solution because the original equation isn't defined if x = 0.
[tex]\sqrt{1 + x} - \sqrt{1 - x}~=~1[/tex]

With equations involving radicals like these, it's a little quicker to move one of them to one side of the equation before squaring both sides.

[tex]\sqrt{1 + x}~=~\sqrt{1 - x} + 1[/tex]

Now square both sides, do some simplification, and square both sides once more.

I end up with x = +/-sqrt(3)/2
 
  • #9
Mark44 said:
[tex]\frac{1}{\sqrt{1 - x} + 1} + \frac{1}{\sqrt{1 + x} - 1}~=~\frac{1}{x}[/tex]

If you do exactly as I said in post 5, this equation becomes
[tex]\frac{-\sqrt{1 - x} + 1}{x} + \frac{\sqrt{1 + x} - 1}{x}~=~\frac{1}{x}[/tex]

Now multiply both sides by x. This won't cause us to lose a solution because the original equation isn't defined if x = 0.
[tex]\sqrt{1 + x} - \sqrt{1 - x}~=~1[/tex]

With equations involving radicals like these, it's a little quicker to move one of them to one side of the equation before squaring both sides.

[tex]\sqrt{1 + x}~=~\sqrt{1 - x} + 1[/tex]

Now square both sides, do some simplification, and square both sides once more.

I end up with x = +/-sqrt(3)/2

Ahhh. That's nice. I see where I went wrong. When I was multiplying each term on the LHS by 1, I didn't change the sign in front of the 1. In other words, I multiplied the first term by sqrt(1 - x) + 1 over itself and the second term by sqrt(1 + x) - 1 over itself. That is a nice tool. I am going to keep that in my pocket for later. Thanks for showing us that.
 
  • #10
The basic idea here is a number times its conjugate being used to simplify an expression with a square root. (a + b)*(a - b) = a^2 - b^2. If a and b are square root expressions, this gets rid of them. Note that the same trick doesn't work for cube roots, but there is another one that might be helpful, namely (a + b)(a^2 - ab + b^2) = a^3 + b^3. Also, as appropriate, (a - b)(a^2 + ab + b^2) = a^3 - b^3.
 

Related to Solving Quadratic Equation: 1{(1-x)^1/2+1} +1{(1+x)^1/2-1}=1/x

What is a quadratic equation?

A quadratic equation is a mathematical expression that contains a variable raised to the second power (x^2) and can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.

How do you solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2-4ac)) / 2a. Alternatively, you can also factor the equation or use the square root property.

What are the roots of a quadratic equation?

The roots of a quadratic equation are the values of x that make the equation true. They are also known as the solutions or zeros of the equation.

Can a quadratic equation have more than two roots?

No, a quadratic equation can only have two roots. This is because a quadratic equation is a second-degree polynomial and the Fundamental Theorem of Algebra states that a polynomial of degree n can have at most n distinct roots.

What is the importance of solving quadratic equations?

Solving quadratic equations is important in various fields of science and mathematics, such as physics, engineering, and economics. It allows us to find the solution to real-world problems and make predictions based on mathematical models.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
646
  • Precalculus Mathematics Homework Help
Replies
5
Views
823
  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
9
Views
935
  • Precalculus Mathematics Homework Help
Replies
2
Views
943
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
Back
Top