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anonymous
 one year ago
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
 one year ago
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

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steve816
 one year ago
Best ResponseYou've already chosen the best response.1What you want to do first is to get the x alone. I'll show you!dw:1440177092884:dw

steve816
 one year ago
Best ResponseYou've already chosen the best response.1Now we take the square root on both sides of the equation.dw:1440177139414:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry just realized I put up the wrong question, I already solved that one. @steve816

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve the equation by factoring. 5z^2  21z + 4 = 0

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0wat's the answer u got ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the first question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need help with that one

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0\[5z ^{2}21z+4=0\]\[5z ^{2}20z1z+4=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why did u take out the one from 21

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0it'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 20z1z= 21z right? So we substitute those 2 numbers instead of 21z !

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure if u got this !

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0\[5z ^{2}20z1z+4=0\] \[5z(z4)1(z4)=0\] did u get what I did in this step ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, if we multiply the numbers we get \[5z^2 20z 1z + 4 = 0\]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0yeah since we have two (z4) Now we gonna consider only one of it ! So now:\[5z(z4)1z(z4)=0\]\[(5z1) (z4)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand this.. I have a doctor appointment now, if u can help me later on.. sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need help with this question.. I have a lot more to do :P

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0first you need to get `like` terms on one side and variable x on the other side

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0so move the 15 to the right how would you do that >

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0no it's already negative at left side to cancel out you would do `opposite` of `subtraction`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to add.. x^2 = 34 + 15?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0and then take square root `both` sides to cancel out square of x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we don't solve for 34 + 15

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0we should solve for x (i guess it supposed to be `z`)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0you answer choices are wrong....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait Lol the question up there is the wrong question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I put it up there that it was wrong.. sorry just realized we were doing the wrong Q

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve the equation by factoring. 5z^2  21z + 4 = 0

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0there are like 4 method to solve quadratic equations you can use the quadratic formula or you can find factors of that equation

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0to 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 .

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0hmm nope dw:1440193043766:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0no 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't really get it, what am I supposed to do now?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0what two number would you multiply to get 20 ? but when you add them you should get 21

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when u add them all up u get 21

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes but \[4+5 \cancel{=} 20\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0you 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 ab) to get middle term

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0what other two numbers you can multiply to get 20 ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0perfect 20 times 1 =20 20+1=21

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm 5z^2+1z+20z+4\] now use group method to factor it

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440193728090:dw make a group of two terms what is common in first two terms ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we add those with the z first?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0no if you add them you will get original equation 5z^2+20x+4 :P

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0find common of `5x^2+1x` and common factor of `20x+4`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay for 5x^2 + 1x its 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then 2 for 20x + 4

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0not just one what about the variable ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0x is the common factor?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0factors of 20 are: `20,5,4,2,1` factors of 4 are : `4,2,1` which one is greatest `common` factor ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440194265638:dw when you take out 1z from `5z^2+1x` what will have left in the parentheses

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0remember both parentheses should be the same!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0what will get get when you take out z from 5z^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05^2 and then if u solve it it will be 10

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0in other words divide both terms by common factor

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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
 one year ago
Best ResponseYou've already chosen the best response.0(5^2 + 1) + (5z + 1)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0don't forget the common factor that would stay outside the parentheses do you mean 5 to the 2 power ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times z }{ z}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Umm so we remove the 2 and not the z or both?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2 \times 2 }{ 2}\] would you cancel both 2s's or just one ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \heartsuit \times \cancel{\heartsuit}}{\cancel{ \heartsuit }}\] one heart canceled with one heart.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0just 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times\cancel{ z} }{\cancel{ z}}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0now both parentheses are common so you can take it out

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440195435458:dw when you take out (5z+1) from z(5z+1) what will get