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

- anonymous

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- anonymous

@steve816

- steve816

What you want to do first is to get the x alone. I'll show you!|dw:1440177092884:dw|

- steve816

Now we take the square root on both sides of the equation.|dw:1440177139414:dw|

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## More answers

- anonymous

sorry just realized I put up the wrong question, I already solved that one. @steve816

- anonymous

5z^2 - 21z + 4 = 0

- anonymous

Solve the equation by factoring. 5z^2 - 21z + 4 = 0

- Rushwr

wat's the answer u got ?

- anonymous

for the first question?

- anonymous

\[\pm7\]

- Rushwr

the 2nd one

- anonymous

I need help with that one

- Rushwr

\[5z ^{2}-21z+4=0\]

- Rushwr

\[5z ^{2}-21z+4=0\]\[5z ^{2}-20z-1z+4=0\]

- Rushwr

did u get tht part?

- anonymous

why did u take out the one from 21

- 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

I'm not sure if u got this !

- anonymous

yeah I kinda do

- Rushwr

sure?

- Rushwr

\[5z ^{2}-20z-1z+4=0\]
\[5z(z-4)-1(z-4)=0\] did u get what I did in this step ?

- anonymous

yes, if we multiply the numbers we get \[5z^2 -20z -1z + 4 = 0\]

- 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

I don't understand this.. I have a doctor appointment now, if u can help me later on.. sorry

- Rushwr

oki

- Nnesha

i like ur pfp. :)

- anonymous

I need help with this question.. I have a lot more to do :P

- Nnesha

first you need to get `like` terms on one side and variable x on the other side

- Nnesha

so move the -15 to the right how would you do that >

- Nnesha

?*

- anonymous

x^2 = 34 - 15?

- Nnesha

no it's already negative at left side
to cancel out you would do `opposite` of `subtraction`

- anonymous

to add.. x^2 = 34 + 15?

- Nnesha

yes right

- Nnesha

and then take square root `both` sides to cancel out square of x.

- anonymous

so we don't solve for 34 + 15

- Nnesha

we should solve for x (i guess it supposed to be `z`)

- Nnesha

you answer choices are wrong....

- Nnesha

your*

- anonymous

wait Lol the question up there is the wrong question

- Nnesha

ye...

- anonymous

I put it up there that it was wrong.. sorry just realized we were doing the wrong Q

- anonymous

Solve the equation by factoring. 5z^2 - 21z + 4 = 0

- 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

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

21z + 4 ?

- Nnesha

hmm nope |dw:1440193043766:dw|

- anonymous

Ooh A

- anonymous

5z^2 + 4

- 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

I don't really get it, what am I supposed to do now?

- Nnesha

what two number would you multiply to get 20 ? but when you add them you should get 21

- anonymous

5*4 = 20
2*10 = 20

- anonymous

when u add them all up u get 21

- Nnesha

yes
but \[4+5 \cancel{=} 20\]

- Nnesha

no just 2 number

- 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

what other two numbers you can multiply to get 20 ?

- anonymous

2 and 10

- Nnesha

and??

- anonymous

1 and 20.

- Nnesha

perfect
20 times 1 =20
20+1=21

- anonymous

Ohh I see

- Nnesha

\[\huge\rm 5z^2+1z+20z+4\] now use group method to factor it

- Nnesha

|dw:1440193728090:dw| make a group of two terms
what is common in first two terms ?

- anonymous

so we add those with the z first?

- Nnesha

no if you add them you will get original equation
5z^2+20x+4 :P

- Nnesha

find common of `5x^2+1x` and common factor of `20x+4`

- anonymous

okay for 5x^2 + 1x its 1?

- anonymous

and then 2 for 20x + 4

- Nnesha

not just one what about the variable ?

- 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

x is the common factor?

- Nnesha

factors of 20 are: `20,5,4,2,1`
factors of 4 are : `4,2,1`
which one is greatest `common` factor ?

- anonymous

4

- Nnesha

yes right

- Nnesha

|dw:1440194265638:dw|
when you take out 1z from `5z^2+1x` what will have left in the parentheses

- Nnesha

remember both parentheses should be the same!

- Nnesha

`5z^2+1z` not x :P

- anonymous

5z^2

- Nnesha

what will get get when you take out z from 5z^2 ?

- anonymous

5^2 and then if u solve it it will be 10

- Nnesha

in other words divide both terms by common factor

- 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

(5^2 + 1) + (5z + 1)

- Nnesha

don't forget the common factor that would stay outside the parentheses
do you mean 5 to the 2 power ?

- anonymous

yes

- Nnesha

\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times z }{ z}\]

- Nnesha

can you cancel out z?

- anonymous

Umm so we remove the 2 and not the z or both?

- Nnesha

\[\frac{ 2 \times 2 }{ 2}\] would you cancel both 2s's or just one ?

- anonymous

I'm not sure

- Nnesha

\[\frac{ \heartsuit \times \cancel{\heartsuit}}{\cancel{ \heartsuit }}\]
one heart canceled with one heart.

- 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

\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times\cancel{ z} }{\cancel{ z}}\]

- anonymous

oh okay

- Nnesha

now both parentheses are common so you can take it out

- Nnesha

wait there is a typo

- 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

|dw:1440195435458:dw|
when you take out (5z+1) from z(5z+1) what will get

- anonymous

z .. I'm kinda confused

- 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

(5z + 1)

- Nnesha

yes right take it out

- Nnesha

just like we did before .

- anonymous

take both of them out?

- Nnesha