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 tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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