Homework Help: Solution to quadratic equation

1. Jun 30, 2016

paulmdrdo

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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

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 Im 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.

2. Jun 30, 2016

ehild

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. Jun 30, 2016

paulmdrdo

The one and only posibility that √(x−q)+√(x−p) =0 is if x=p=q. Therefore x=q=p is also a solution? Im still confused. Please bear with me.

4. Jun 30, 2016

Ray Vickson

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. Jun 30, 2016

paulmdrdo

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. Jun 30, 2016

SammyS

Staff Emeritus
The following doesn't make sense unless p = 0 .

7. Jun 30, 2016

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.

Last edited: Jun 30, 2016
8. Jul 1, 2016

ehild

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. Jul 1, 2016

Ray Vickson

$$\text{original right-hand-side} = \frac{q}{\sqrt{x-p}} + \frac{p}{\sqrt{x-q}}$$
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
$$\text{new right-hand-side} =q \sqrt{x-q} + p \sqrt{x-p}$$
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.

Last edited: Jul 1, 2016
10. Jul 1, 2016

paulmdrdo

Thank you now it is clear!