Solution to quadratic equation

In summary, the conversation revolved around solving a quadratic equation and the potential issue of discarding one of the roots while dividing by a factor containing x. It was clarified that this factor can only be equal to zero if x equals both p and q, making it an invalid solution. However, after multiplying the equation by the factor, it is possible to have x equal to p or q in the new equation.
  • #1
paulmdrdo
89
2
Q48_49hall_XXVIC.jpg
1. Homework Statement

Homework Equations

The Attempt at a Solution



I just want some clarification about the rules on solving quadratic equation. My question is coming from my solution for problem 49.

Solving problem 49 [/B]
Factor: (x−p−q)(√(x−q)+√(x−p))=0
Therefore: x−p−q=0⇒x=p+q this agrees with the answer in my book

What I am confused about is the part where I divided out the lhs by (√(x−q)+√(x−p)). This factor contains x in it and I'm worried that I may have discarded the other root for the equation. And since it is a quadratic equation I am expecting to get two roots for the equation. Can you enlighten me about this matter? Thanks.
 
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  • #2
paulmdrdo said:
Q48_49hall_XXVIC.jpg
1. Homework Statement

Homework Equations

The Attempt at a Solution



I just want some clarification about the rules on solving quadratic equation. My question is coming from my solution for problem 49.

Solving problem 49 [/B]
Factor: (x−p−q)(√(x−q)+√(x−p))=0
Therefore: x−p−q=0⇒x=p+q this agrees with the answer in my book

What I am confused about is the part where I divided out the lhs by (√(x−q)+√(x−p)). This factor contains x in it and I'm worried that I may have discarded the other root for the equation. And since it is a quadratic equation I am expecting to get two roots for the equation. Can you enlighten me about this matter? Thanks.
When you solve an equation of form f(x) g(x) =0, you use the fact that the product is zero if either f(x)=0 or g(x)=0. In principle, it is possible that √(x−q)+√(x−p) = 0. What does it mean for √(x−q) and √(x−p)?
You started the solution by multiplying the whole equation with √(x−q)√(x−p). What is the condition that you may do it?
 
  • #3
The one and only posibility that √(x−q)+√(x−p) =0 is if x=p=q. Therefore x=q=p is also a solution? I am still confused. Please bear with me.
 
  • #4
paulmdrdo said:
The one and only posibility that √(x−q)+√(x−p) =0 is if x=p=q. Therefore x=q=p is also a solution? I am still confused. Please bear with me.

You are not allowed to have either ##x = p## or ##x = q##, because the right-hand-side of your original equation would then contain division by zero, which is absolutely forbidden.
 
  • #5
Ray Vickson said:
You are not allowed to have either ##x = p## or ##x = q##, because the right-hand-side of your original equation would then contain division by zero, which is absolutely forbidden.

What? When x = x+p we will also have division by zero right? Why is it that √(x-q)+√(x-p) is not allowed to be equal zero but x-p-q can be equal zero?
 
  • #6
paulmdrdo said:
What? When x = x+p we will also have division by zero right? Why is it that √(x-q)+√(x-p) is not allowed to be equal zero but x-p-q can be equal zero?
The following doesn't make sense unless p = 0 .
paulmdrdo said:
When x = x+p
 
  • #7
Suggestion: Group the sqrt (x-p ) terms on one side of the equation (one term will be in the denominator), and the sqrt(x-q) terms on the other side of the equation and square both sides of the equation. You can then do some additional algebra. It's then a couple more steps, but you do get a quadratic expression for x in terms of p and q. The quadratic expression does factor and simplify and it does have a simple solution.
 
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  • #8
paulmdrdo said:
The one and only posibility that √(x−q)+√(x−p) =0 is if x=p=q. Therefore x=q=p is also a solution? I am still confused. Please bear with me.
You started the solution with multiplying the whole equation by √(x−q)*√(x−p). Before multiplying an equation with an expression you have to exclude that it is zero.
 
  • #9
paulmdrdo said:
What? When x = x+p we will also have division by zero right? Why is it that √(x-q)+√(x-p) is not allowed to be equal zero but x-p-q can be equal zero?

Please read responses carefully: I said "... original equation...". In your original equation you had
[tex] \text{original right-hand-side} = \frac{q}{\sqrt{x-p}} + \frac{p}{\sqrt{x-q}} [/tex]
If you take ##x = p## or ##x = q## you will have ##1/\sqrt{0}## on the right, and that is forbidden.

Of course, after you multiply your equation by ##\sqrt{x-p} \sqrt{x-q}## you would have
[tex] \text{new right-hand-side} =q \sqrt{x-q} + p \sqrt{x-p} [/tex]
and in that new rhs you are certainly allowed to have ##x = p## or ##x = q##. However, that would not be giving you a solution to the original equation.
 
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  • #10
Ray Vickson said:
Please read responses carefully: I said "... original equation...". In your original equation you had
[tex] \text{original right-hand-side} = \frac{1}{\sqrt{x-p}} + \frac{1}{\sqrt{x-q}} [/tex]
If you take ##x = p## or ##x = q## you will have ##1/\sqrt{0}## on the right, and that is forbidden.

Of course, after you multiply your equation by ##\sqrt{x-p} \sqrt{x-q}## you would have
[tex] \text{new right-hand-side} = \sqrt{x-q} + \sqrt{x-p} [/tex]
and in that new rhs you are certainly allowed to have ##x = p## or ##x = q##. However, that would not be giving you a solution to the original equation.
Thank you now it is clear!
 

1. What is a quadratic equation?

A quadratic equation is a mathematical expression that contains a variable raised to the power of two. It is typically written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

2. 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 factor the equation or complete the square to find the roots.

3. What are the solutions to a quadratic equation?

The solutions to a quadratic equation are the values of x that make the equation true. There can be two real solutions, two complex solutions, or one repeated solution.

4. What does the discriminant of a quadratic equation represent?

The discriminant, b^2 - 4ac, is a term used to determine the nature of the solutions to a quadratic equation. If the discriminant is positive, there are two distinct real solutions. If it is zero, there is one repeated solution. And if it is negative, there are two complex solutions.

5. How can quadratic equations be applied in real life?

Quadratic equations have many real-world applications, such as calculating the trajectory of a projectile, determining the maximum or minimum value of a function, and solving optimization problems. They are also commonly used in fields like physics, engineering, and economics.

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