Solve 0=z^2−10z+25 by factoring
@dude @jhonyy9
write the 10 as product of two numbers so that the 25 is the square of one number of these two
can one number be a decimal? and why a square? we didn't learn it as a square
25 is a square of what number ?
5
?
exactly so using the 5 how you get 10 ?
x2
exactly and now rewrite this equation using these above wrote
wait, it's x(-2) bc it's -10x and would it be z^2+5x-2x+25?
no please make what i say
10 = 2*5 25 = 5*5
ya but but the 10 is negative in the equation
and why not can rewrite it in this way -(2*5)
well i didn't know that, we learned it a different way
ok show me please how ?
\[z^2-10x+25=0\] we need to break it into 4 terms by splitting up the -10x to do so we do 1x25=25 and a set of factors of 25 needs to =-10 we learned the "x method" i need to find what factors of 25 = -10 but idk how |dw:1603832942553:dw|
ohhh, yeah i get that way too. she briefly went over that so we didn't really know that method. so if i am trying to find the solution(s) aka x-intercept(s) from the expression, how do i do that?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 \(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 sorry i know different method z^2 -10z +25 = 0 z*z - 5z -5z +5*5 = 0 so from first two terms factorize out the z and will get z(z-5) and from 3rd and 4th term factorize out the -5 and will get -5(z-5) so in this way our equation will be z(z-5) -5(z-5) = 0 but now you see that every two terms contain a common term of (z-5) so what can be factorize it out and in this way will get (z-5)(z-5) = 0 what is a square so can be write in the form of (z-5)^2 = 0 do you understand it now clearly ? \(\color{#0cbb34}{\text{End of Quote}}\) \(\color{#0cbb34}{\text{End of Quote}}\)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 \(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 \(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 sorry i know different method z^2 -10z +25 = 0 z*z - 5z -5z +5*5 = 0 so from first two terms factorize out the z and will get z(z-5) and from 3rd and 4th term factorize out the -5 and will get -5(z-5) so in this way our equation will be z(z-5) -5(z-5) = 0 but now you see that every two terms contain a common term of (z-5) so what can be factorize it out and in this way will get (z-5)(z-5) = 0 what is a square so can be write in the form of (z-5)^2 = 0 do you understand it now clearly ? \(\color{#0cbb34}{\text{End of Quote}}\) \(\color{#0cbb34}{\text{End of Quote}}\) \(\color{#0cbb34}{\text{End of Quote}}\) ya
sure - any questions ?
so if i am trying to find the solution(s) aka x-intercept(s) from the expression, how do i do that? because she said we have to find the solutions by factoring the expression
ok so how is this equation above factorized ?
well i only know it written out the way you did it, and i only know how to find solutions from the way i learned it in school so im confused
please answer my above wrote question - i will teach you how you get the zero(s) of this equation
zero(s) = solutions
(z-5)^2 ?? i'm confused... and when you factorized the equation above^ idk what you did for the 25 because you broke it apart but idk what you did with it after that. so i'm lost, can you reexplain it, sorry
ok do you agree it that (z-5)^2 = z^2 -10z +25 ?
you need to know formula of (a -b)^2 = a^2 -2ab +b^2
let me add that to my notes, i didn't know that
ok so now,i agree that in (z-5)^2 = z^2 but i don't get how the rest comes out to be -10z +25
ok try it please multiplie the (z-5) by (z-5)
than you know this my above wrote formula and use this or you calcule it step by step
ok so you multiply x*x and get x^2 -5*-5 is 25 so if i did that correctly ^, where do we get the -10x from that?
multiplie term by term you forget the z*(-5) and -5(z) what will give you the -10z
why do you do z * -5, it's not written?
(z-5)(z-5) = ? you need multiplie term by term
|dw:1603835097191:dw|
ok ok ok, i get that one, that's easier to understand
first z by second z and first z by second -5 and
this is sure now ?
i understand how got the -10x but i still need to find the zeros
ok so this is easy after you get an equation factorized just make these factor(s) equal by zero and calcule the variable
bc. than you check an equation factorized than from these factori one is equal zero so this equation will give you equal zero - yes ?
i think ?
prove it
in this above wrote case how will be ?
please write the factorized equation
waitttttt, if you have (z-5)^2 (which is equal to z^2-10z+25=0) can you write it out as (z-5)=0 and again so you have \[(z-5)=0 ( z-5)=0\] and since thy're the same you only have to doit once and when you solve for z you get 5 so is the zero of this equation (5,0) ???
5 is correct but 0 why is ?
well we have to write it as an ordered pair because we are looking for the zero(s) of the parabola
but without looking at a graph
i dont know about what you sai but there is just z_1 = 5 zero - solution of this equation 0 not is correct bc. then you substitute the 0 in place of z not will get never equal with zero the equation right ?
so just write it as 5?
and yes
exactly and was my pleasure
thank you !!
anytime bye bye
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