anonymous
  • anonymous
Solve the equation using square roots. x^2– 15 = 34 A) z = 1/5 or z = -4 B) z = 1 or z = -4 C) z = 1/5 or z = 4 D) z = 1 or z = 4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@steve816
steve816
  • steve816
What you want to do first is to get the x alone. I'll show you!|dw:1440177092884:dw|
steve816
  • steve816
Now we take the square root on both sides of the equation.|dw:1440177139414:dw|

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More answers

anonymous
  • anonymous
sorry just realized I put up the wrong question, I already solved that one. @steve816
anonymous
  • anonymous
5z^2 - 21z + 4 = 0
anonymous
  • anonymous
Solve the equation by factoring. 5z^2 - 21z + 4 = 0
Rushwr
  • Rushwr
wat's the answer u got ?
anonymous
  • anonymous
for the first question?
anonymous
  • anonymous
\[\pm7\]
Rushwr
  • Rushwr
the 2nd one
anonymous
  • anonymous
I need help with that one
Rushwr
  • Rushwr
\[5z ^{2}-21z+4=0\]
Rushwr
  • Rushwr
\[5z ^{2}-21z+4=0\]\[5z ^{2}-20z-1z+4=0\]
Rushwr
  • Rushwr
did u get tht part?
anonymous
  • anonymous
why did u take out the one from 21
Rushwr
  • Rushwr
it's like this u see the middle number 21z right? 1st we multiply 4 from 5 which is 20 right? Now we know we need 4 numbers right? So using the factors of 20z we are gonna make 21z 20z* 1 is 20z right? but we also know -20z-1z= 21z right? So we substitute those 2 numbers instead of 21z !
Rushwr
  • Rushwr
I'm not sure if u got this !
anonymous
  • anonymous
yeah I kinda do
Rushwr
  • Rushwr
sure?
Rushwr
  • Rushwr
\[5z ^{2}-20z-1z+4=0\] \[5z(z-4)-1(z-4)=0\] did u get what I did in this step ?
anonymous
  • anonymous
yes, if we multiply the numbers we get \[5z^2 -20z -1z + 4 = 0\]
Rushwr
  • Rushwr
yeah since we have two (z-4) Now we gonna consider only one of it ! So now:\[5z(z-4)-1z(z-4)=0\]\[(5z-1) (z-4)=0\]
anonymous
  • anonymous
I don't understand this.. I have a doctor appointment now, if u can help me later on.. sorry
Rushwr
  • Rushwr
oki
Nnesha
  • Nnesha
i like ur pfp. :)
anonymous
  • anonymous
I need help with this question.. I have a lot more to do :P
Nnesha
  • Nnesha
first you need to get `like` terms on one side and variable x on the other side
Nnesha
  • Nnesha
so move the -15 to the right how would you do that >
Nnesha
  • Nnesha
?*
anonymous
  • anonymous
x^2 = 34 - 15?
Nnesha
  • Nnesha
no it's already negative at left side to cancel out you would do `opposite` of `subtraction`
anonymous
  • anonymous
to add.. x^2 = 34 + 15?
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
and then take square root `both` sides to cancel out square of x.
anonymous
  • anonymous
so we don't solve for 34 + 15
Nnesha
  • Nnesha
we should solve for x (i guess it supposed to be `z`)
Nnesha
  • Nnesha
you answer choices are wrong....
Nnesha
  • Nnesha
your*
anonymous
  • anonymous
wait Lol the question up there is the wrong question
Nnesha
  • Nnesha
ye...
anonymous
  • anonymous
I put it up there that it was wrong.. sorry just realized we were doing the wrong Q
anonymous
  • anonymous
Solve the equation by factoring. 5z^2 - 21z + 4 = 0
Nnesha
  • Nnesha
there are like 4 method to solve quadratic equations you can use the quadratic formula or you can find factors of that equation
Nnesha
  • Nnesha
to factor it find two number when you multiply them you should get product of AC and when you add them you should get middle term \[\huge\rm Ax^2+Bx+C=0\] a=leading coefficient b=middle term c=constant term so what is A and C in that equation .
anonymous
  • anonymous
21z + 4 ?
Nnesha
  • Nnesha
hmm nope |dw:1440193043766:dw|
anonymous
  • anonymous
Ooh A
anonymous
  • anonymous
5z^2 + 4
Nnesha
  • Nnesha
no just the coefficient so A = 5 b=21 c=4 now multiply a times c = 5 times 4 = 20 find two number when you multiply them you should get 20 but when you add or subtract them you should get middle term which is 21
anonymous
  • anonymous
I don't really get it, what am I supposed to do now?
Nnesha
  • Nnesha
what two number would you multiply to get 20 ? but when you add them you should get 21
anonymous
  • anonymous
5*4 = 20 2*10 = 20
anonymous
  • anonymous
when u add them all up u get 21
Nnesha
  • Nnesha
yes but \[4+5 \cancel{=} 20\]
Nnesha
  • Nnesha
no just 2 number
Nnesha
  • Nnesha
you can't add 4 number to get 20 here is an example let's say a and b are numbers so when i multiply them a times b = 20 i should add `same` two numbers a+b (or a-b) to get middle term
Nnesha
  • Nnesha
what other two numbers you can multiply to get 20 ?
anonymous
  • anonymous
2 and 10
Nnesha
  • Nnesha
and??
anonymous
  • anonymous
1 and 20.
Nnesha
  • Nnesha
perfect 20 times 1 =20 20+1=21
anonymous
  • anonymous
Ohh I see
Nnesha
  • Nnesha
\[\huge\rm 5z^2+1z+20z+4\] now use group method to factor it
Nnesha
  • Nnesha
|dw:1440193728090:dw| make a group of two terms what is common in first two terms ?
anonymous
  • anonymous
so we add those with the z first?
Nnesha
  • Nnesha
no if you add them you will get original equation 5z^2+20x+4 :P
Nnesha
  • Nnesha
find common of `5x^2+1x` and common factor of `20x+4`
anonymous
  • anonymous
okay for 5x^2 + 1x its 1?
anonymous
  • anonymous
and then 2 for 20x + 4
Nnesha
  • Nnesha
not just one what about the variable ?
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @proud_yemeniah_ and then 2 for 20x + 4 \(\color{blue}{\text{End of Quote}}\) greatest common factor yes two is common but it's not greatest common
anonymous
  • anonymous
x is the common factor?
Nnesha
  • Nnesha
factors of 20 are: `20,5,4,2,1` factors of 4 are : `4,2,1` which one is greatest `common` factor ?
anonymous
  • anonymous
4
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
|dw:1440194265638:dw| when you take out 1z from `5z^2+1x` what will have left in the parentheses
Nnesha
  • Nnesha
remember both parentheses should be the same!
Nnesha
  • Nnesha
`5z^2+1z` not x :P
anonymous
  • anonymous
5z^2
Nnesha
  • Nnesha
what will get get when you take out z from 5z^2 ?
anonymous
  • anonymous
5^2 and then if u solve it it will be 10
Nnesha
  • Nnesha
in other words divide both terms by common factor
Nnesha
  • Nnesha
hmm no \[\huge\rm \color{Red}{\frac{ 5z^2 }{ z }+\frac{ 1z }{ z }}+(\frac{ 20z }{ 4 }+\frac{ 4 }{ 4 })\] \[\huge\rm \color{Red}{z(??+??)}+4(??+??)\] divide by common and then write your answer in the parentheses
anonymous
  • anonymous
(5^2 + 1) + (5z + 1)
Nnesha
  • Nnesha
don't forget the common factor that would stay outside the parentheses do you mean 5 to the 2 power ?
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times z }{ z}\]
Nnesha
  • Nnesha
can you cancel out z?
anonymous
  • anonymous
Umm so we remove the 2 and not the z or both?
Nnesha
  • Nnesha
\[\frac{ 2 \times 2 }{ 2}\] would you cancel both 2s's or just one ?
anonymous
  • anonymous
I'm not sure
Nnesha
  • Nnesha
\[\frac{ \heartsuit \times \cancel{\heartsuit}}{\cancel{ \heartsuit }}\] one heart canceled with one heart.
Nnesha
  • Nnesha
just one \[\frac{ 2 \times \cancel{2} }{ \cancel{2}}\] you can multiply 2 at the top you will get 4/2 when u divide them you will get 2 if you cancel out both 2 with one you will get `1` as an answer which is wrong
Nnesha
  • Nnesha
\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times\cancel{ z} }{\cancel{ z}}\]
anonymous
  • anonymous
oh okay
Nnesha
  • Nnesha
now both parentheses are common so you can take it out
Nnesha
  • Nnesha
wait there is a typo
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @proud_yemeniah_ (5^2 + 1) + (5z + 1) \(\color{blue}{\text{End of Quote}}\) so it should \[\huge\rm z(5z+1)+4(5z+1)\] and the hing i gave you both parentheses should be the same
Nnesha
  • Nnesha
|dw:1440195435458:dw| when you take out (5z+1) from z(5z+1) what will get
anonymous
  • anonymous
z .. I'm kinda confused
Nnesha
  • Nnesha
okay \[\huge\rm \color{Red}{z(5z+1)}+\color{blue}{4(5z+1)}\] red= first term blue=2nd term now what is common in both terms ?
anonymous
  • anonymous
(5z + 1)
Nnesha
  • Nnesha
yes right take it out
Nnesha
  • Nnesha
just like we did before .
anonymous
  • anonymous
take both of them out?
Nnesha
  • Nnesha
Yes right!
Nnesha
  • Nnesha
\[(5z+1)(??+??)\] when you take out the common factor what will u write in the parentheses :)
anonymous
  • anonymous
I really don't know what to do now..
Nnesha
  • Nnesha
what will you get when you take out the 5x+1 from z(5z+1 )
anonymous
  • anonymous
1?
Nnesha
  • Nnesha
lets say 5z+1 = chocolates and z = apple when you take out chocolates from apple what will you get
anonymous
  • anonymous
U can't do that
Nnesha
  • Nnesha
hmm why not ?
anonymous
  • anonymous
bc apples are different from chocolate
Nnesha
  • Nnesha
\[\huge\rm \color{Red}{apples(chocolates)}+\color{blue}{cookies(chocolates)}\] blue = 1st plate red=2nd plate okay ? so you take out chocolates from red plate what will you have left in red plate ?
anonymous
  • anonymous
apples which is 5z?
Nnesha
  • Nnesha
no apples is z
Nnesha
  • Nnesha
apples = z (5z+1)= chocolates
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
so we now have z left
Nnesha
  • Nnesha
yes right!!!
Nnesha
  • Nnesha
what about blue plate ?
anonymous
  • anonymous
4(5z + 1)?
Nnesha
  • Nnesha
that's a plate where we can say 4 represents cookies and 5z+1 is same (chocolates) i''m saying take out the choclates from both plate and give it to me bec i love them
anonymous
  • anonymous
so 4 is left
Nnesha
  • Nnesha
yes right when you took chocolates from red plate u got apples and when you take out chocolates from blue plate you hve left with cookies \[\huge\rm chocolates (apples +cookies)\] `chocolates` `5z+1` `apples` = `z` `cookies`= `4` \[(5z+1)(\color{Red}{z}+\color{blue}{4})\]
Nnesha
  • Nnesha
set both parentheses equal to zero \[(5z+1)=0~~~(\color{Red}{z}+\color{blue}{4})=0\]solve for z
anonymous
  • anonymous
so its just like we would solve for x we put any number for z ?
Nnesha
  • Nnesha
hmm ye that would work too you can use answer choices plug in x value if you get equal sides then that would be your answer or just solve for z z+4 =0subtract 4 both sides
anonymous
  • anonymous
Can u help me to understand this more
Nnesha
  • Nnesha
sure. so 5z+1=0 solve for 5 move the 1 and 5 to the right side
Nnesha
  • Nnesha
how would you cancel out 1 from left side ?
anonymous
  • anonymous
you add them or what?
Nnesha
  • Nnesha
no.
Nnesha
  • Nnesha
not add it's already positive to cancel you should apply opposite of addition 1+1 =2 not 0
Nnesha
  • Nnesha
what is opposite of subtraction ?
anonymous
  • anonymous
to add
Nnesha
  • Nnesha
yes and opposite of add is to subtract so subtract 1 both sides
anonymous
  • anonymous
so we also subtract 1 from the 4
Nnesha
  • Nnesha
no
Nnesha
  • Nnesha
\[5z+1=0\] first should solve this one for z
Nnesha
  • Nnesha
there isn't any 4
Nnesha
  • Nnesha
you would get 2 solutions(2 z's values ) in quadratic equation always!
Nnesha
  • Nnesha
first solve 5z+1=0 for z and then solve z+4 =0 for z
anonymous
  • anonymous
so 4 for 5z + 1
Nnesha
  • Nnesha
no no remember we got 2 parnetheses first one is `(5z+1)` and the 2nd one is `z+4` set both equal to zero `5z+1=0` and `z+4=0`
Nnesha
  • Nnesha
gawd hmm i just realized it's -21z not 21z :(
Nnesha
  • Nnesha
good think is we just have to change sign -20 times -1 = 20 -20-1=-21 \[\huge\rm 5z^2-z-20z+4\] \[z(5z-1)-4(5z-1)\] \[(z-4)(5z-1)=0\]
Nnesha
  • Nnesha
\[\huge\rm z-4=0\] solve for z add 4 both sides
Nnesha
  • Nnesha
\[\huge\rm 5z-1=0\] solve for z 1) add 1 both sides 2) divide both sides by 5
Nnesha
  • Nnesha
gtg bye.

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