Embarassed to ask this but 0 = x2 – 15x + 50 ?

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The discussion revolves around solving the quadratic equation 0 = x² – 15x + 50 by factoring. The correct factorization is identified as (x - 5)(x - 10) = 0. Participants clarify the steps involved in factoring, emphasizing the importance of recognizing common factors. Misunderstandings about the algebraic manipulation are addressed, particularly regarding incorrect expressions. The conversation concludes with a confirmation of the correct factorization method.
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[SOLVED] embarassed to ask this...but 0 = x2 – 15x + 50 ??

Homework Statement



0 = x2 – 15x + 50

Homework Equations





The Attempt at a Solution



line1 0 = x2 – 15x + 50
line2 0 = (x – 5)(x – 10)

question: how does the equation in line one work out to the equation in the second line?
reviewing my algebra...i can't believe how much high school math I forgot...
 
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x(x - 10) - 5(x - 10)
 
What is your question? You're factoring a binomial.

So, (x+a)(x+b)=0 factors to x*x + a*x + b*x + a*b = 0
 
EnumaElish said:
x(x - 10) - 5(x - 10)
Isn't this
x^2 - 5x + 90
?
 
DaveC426913 said:
Isn't this
x^2 - 5x + 90
?


x(x - 10) - 5(x - 10)

no no it isnt

if you factor out (x-10 from above...you will get (x-10)(x-5)
 
got it...thankx
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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