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Mathematics 65 Online
mxddi3:

Solve 0=z^2−10z+25 by factoring

mxddi3:

@dude @jhonyy9

jhonyy9:

write the 10 as product of two numbers so that the 25 is the square of one number of these two

mxddi3:

can one number be a decimal? and why a square? we didn't learn it as a square

jhonyy9:

25 is a square of what number ?

mxddi3:

5

mxddi3:

?

jhonyy9:

exactly so using the 5 how you get 10 ?

mxddi3:

x2

jhonyy9:

exactly and now rewrite this equation using these above wrote

mxddi3:

wait, it's x(-2) bc it's -10x and would it be z^2+5x-2x+25?

jhonyy9:

no please make what i say

jhonyy9:

10 = 2*5 25 = 5*5

mxddi3:

ya but but the 10 is negative in the equation

jhonyy9:

and why not can rewrite it in this way -(2*5)

mxddi3:

well i didn't know that, we learned it a different way

jhonyy9:

ok show me please how ?

mxddi3:

\[z^2-10x+25=0\] 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|

mxddi3:

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?

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 ? \(\color{#0cbb34}{\text{End of Quote}}\) \(\color{#0cbb34}{\text{End of Quote}}\)

mxddi3:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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

jhonyy9:

sure - any questions ?

mxddi3:

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

jhonyy9:

ok so how is this equation above factorized ?

mxddi3:

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

jhonyy9:

please answer my above wrote question - i will teach you how you get the zero(s) of this equation

jhonyy9:

zero(s) = solutions

mxddi3:

(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

jhonyy9:

ok do you agree it that (z-5)^2 = z^2 -10z +25 ?

jhonyy9:

you need to know formula of (a -b)^2 = a^2 -2ab +b^2

mxddi3:

let me add that to my notes, i didn't know that

mxddi3:

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

jhonyy9:

ok try it please multiplie the (z-5) by (z-5)

jhonyy9:

than you know this my above wrote formula and use this or you calcule it step by step

mxddi3:

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?

jhonyy9:

multiplie term by term you forget the z*(-5) and -5(z) what will give you the -10z

mxddi3:

why do you do z * -5, it's not written?

jhonyy9:

(z-5)(z-5) = ? you need multiplie term by term

jhonyy9:

|dw:1603835097191:dw|

mxddi3:

ok ok ok, i get that one, that's easier to understand

jhonyy9:

first z by second z and first z by second -5 and

jhonyy9:

this is sure now ?

mxddi3:

i understand how got the -10x but i still need to find the zeros

jhonyy9:

ok so this is easy after you get an equation factorized just make these factor(s) equal by zero and calcule the variable

jhonyy9:

bc. than you check an equation factorized than from these factori one is equal zero so this equation will give you equal zero - yes ?

mxddi3:

i think ?

jhonyy9:

prove it

jhonyy9:

in this above wrote case how will be ?

jhonyy9:

please write the factorized equation

mxddi3:

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) ???

jhonyy9:

5 is correct but 0 why is ?

mxddi3:

well we have to write it as an ordered pair because we are looking for the zero(s) of the parabola

mxddi3:

but without looking at a graph

jhonyy9:

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 ?

mxddi3:

so just write it as 5?

mxddi3:

and yes

jhonyy9:

exactly and was my pleasure

mxddi3:

thank you !!

jhonyy9:

anytime bye bye

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