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anonymous
 5 years ago
using quadratic equation how would i solve x^2+13x+22=0 ?
anonymous
 5 years ago
using quadratic equation how would i solve x^2+13x+22=0 ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know what a,b and c are?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my guess is that a = x^2 b= 13x and c= 22 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a is the number in from of the x^2 = 1 b is the number in from of the x = 13 c is the constant = 22 Do you know the quadratic formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i do b plus or minus square root radical b squared minus 4ac all over 2a ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good.. so you put those in (I like to read it the opposite of b), it helps when you put in values.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.013 + sqrt(13^2  4(1)(22)) all over 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do the b^2  4ac next

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes which = 81 13 + sqrt(81) all over 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now... is the sqrt (81) an integer "a perfect number"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so x equals 13 and 9 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(81) = 9 so you now have (13 + 9)/2 Your two answers are (13 + 9)/2 = 4/2 = 2 Other answer (13  9)/2 = 22/2 = 11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get it now... omg thank you so much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Be careful when the stuff under the sqrt root sign does not come out "perfectly" then there are more steps to do... If you have another one that is different.. just post and I'll help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have like 3 more but im going to try them on my own first and if i get stuck ill post it (: thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so like you said i theres more steps for ones that are not "perfectly" well i got 7+ sqrt 33/ 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know 33 is not a perfect square so what would i do next?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0factor out any "perfect" squares from 33.....there aint none.. so leave it as is

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if you got sqrt(33) as an answer, then the only solution available is to go to complex numbers; but usually they just want to know if there are any "real" numbers as a solution. sqrt(33) has no "real" number solutions because any 2 numbers when multiplied together will give you a (+)positive result

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thank you (: so i need help with another problem please? It says to solve by factoring or using quadratic equation; the problem given is (x6) (5x+9) = 0 how would i do this?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok... solve by factoring OR quad formula right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since the equation is ALREADY in a factored form; we use the rule that anything times zero = 0. that means that when (x6) = 0 we have a winner; OR when (5x+9) = 0 we have another winner. So we can get 2 solutions to it, can you tell me what x has to equal to get us a zero?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your "numerals" are correct, but check the "signs" for me.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0exactly, that was very good :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks im trying...i was gone from school alot because my parents are going through divorce and its been causing me to get very sick so i've been trying to get caught up on make up work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for your help though!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0youre quite welcome; and I hope you find better days ahead.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That problem you were doing with 33, can you post the actual problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b^2  4ac = 7^2  4(1)(18) = 7^2 = 49 4(1)(18) = +72 So 49 + 72 = 121 You will find this is a "perfect square root  no decimals" Let me know what you get after this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No... think of it as a (4) times (a)(c) if you do it your way then you have to think of it as 49  (72) which then you have to turn into a + sign... much easier if you use a (4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oooo okayy thank youu!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me know what you get for answers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or do i add 49 and 72 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.07^2  4(1)(18) 49 + 72 = 121 Look at the  between the 7^2 and the 4 we are basically going to turn that into a + and put the  with the 4 so 7^2 + (4)(1)(18) = 49+72 = 121

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then its going to be 7+ 11/2 and then 7 11/2??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes.. do the (7 + 11) first then /2 and (7  11) then /2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answers are 2 and 9 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes... Do you have another one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you! yes i have one more a^23a = 6 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, what is the a,b, and c Remember.... when it says to solve and you have like an x^2, IT MUST EQUAL ZERO FIRST BEFORE YOU FIND YOUR A,B, and C This one does, I am just making sure you know that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is it going to be 3 + sqrt 924/ 2 ? 924 is 15 which isn't perfect so what would i do next ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(before I check that... you had an equal sign before the 6, is it a plus or a minus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, remember a,b and c are the numbers in front of the x^2, the x and the number... KEEP the SIGNS of THOSE NUMBERS with the NUMBERS. a = 1 b = 3 c = 6 Are you in algebra I or II? Have you heard of imaginary numbers yet? i's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0algebra 1 and yes a little bit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am surprised that this is your next problem.. If you are sure you entered it correctly. 3 + sqrt(924)/2 3 + sqrt(15)/2 When you have a "" underneath the sqrt sign you get rid of it by bringing out an "i" in front of the sqrt sign. Notice it is 3 at the beginning because b was 3 3 + i sqrt(15) all over 2 3  i sqrt(15) all over 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im lost with the "i" and ""

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure your problem was correct? I doubt you have learned about imaginary numbers. that is why I think your problem may be incorrect.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no the problem is correct its an extra credit problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would leave it as 15 under the square root... 3 + sqrt(15) /2 3  sqrt(15) /2 Hey.. I read about your family situation. I am a teacher in a school.. I want you to know that Jesus knows what you are going thru and he wants to be there for you. He loves you more than anyone you know. If you ever want me to help you with math, or just "talk", you can email me at amcbeth328@gmail.com My name is Andrea McBeth. Keep working hard. I am very pleased to see that you are trying to learn this.. That is commendable!!! I will be praying for you and your situation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so much mam! god bless <3 i will be sure to add your email to contacts (:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can call me Andrea :) by the way.. I saw another of your posts and the answer was not completely correct. How do you factor 2x^4  32x^2 You factor out the 2x^2(x^2  16) But the (x^2  16) can be factored into (x + 4)(x  4) that is the difference between two squares. So the answer is 2x^2(x + 4)(x  4)
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