Two Problems with Quadratic Equations

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Vertciel

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Hello everyone,

I would appreciate any help for the following two problems. Also, I have shown what I have attempted so far.

Thanks!

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1. Solve the following for x in terms of y:

\(\displaystyle x^2 - 2x + 1 + y^2 = 0\)

\(\displaystyle (x - 1)^2 + y^2 = 0\)

I am not sure how to factor fully this equation or to find the A, B, C, values for using the Quadratic Formula.


2. For a, b, and h real numbers, show that the roots of \(\displaystyle (x - a)(x - b) = h^2\) are always real.

I know that the roots are: \(\displaystyle x = a\) and \(\displaystyle x = b\), and that the discriminant must be equal to or greater than 0 for a quadratic equation to have real roots.
 
Vertciel said:
1. Solve the following for x in terms of y:

\(\displaystyle x^2 - 2x + 1 + y^2 = 0\)

\(\displaystyle (x - 1)^2 + y^2 = 0\)

I am not sure how to factor fully this equation or to find the A, B, C, values for using the Quadratic Formula.
Click to expand...
This is very sad. You know of the quadratic formula but you've no idea where it came from. I blame your teachers, not you.

Revelation: The quadratic formula is developed through the common processof completing the square.

Application: If you already have the x-factor nicely squared, you are almost done and the quadratic formula is slower.

(x - 1)^2 + y^2 = 0
(x - 1)^2 = - y^2
x - 1 = +/- sqrt(- y^2)
x = 1 +/- sqrt(- y^2)

Done!!!

Really important note for this particular problem. I did the algebra above just as an example. We should have stopped this problem right here:

(x - 1)^2 + y^2 = 0

A little thought proves y^2 >= 0. Squaring Real numbers results in Positive values or Zero. The same is the case for (x-1)^2. How can these two things, both greater than or equal to zero, be added and get zero (0)? They BOTH must be zero (0). To hint at some geometry, you may recognize this as the equation of a circle with radius zero (0). It's just a single point! (1,0)
 
Thank you for your help and detailed explanation, tkhunny! I truly appreciate it!

That is what I got also, but the answer at the back of the textbook says that x = 1. Do you know how x = 1 can be gotten?
 
(x - 1)^2 Where is that zero?
y^2 Where is that zero?

It's just a single point! (1,0)
 
Hello tkhunny once again,

Could you please explain your latest post? I still don't completely understand how x = 1.

Thank you very much!
 
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