Question

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

Answer 1

Do you know what a,b and c are?

Answer 2

my guess is that a = x^2 b= 13x and c= 22 ?

Answer 3

a 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?

Answer 4

OOPS "in front"

Answer 5

yes i do -b plus or minus square root radical b squared minus 4ac all over 2a ?

Answer 6

Good.. so you put those in (I like to read it the opposite of b), it helps when you put in values.

Answer 7

-13 +- sqrt(13^2 - 4(1)(22)) all over 2

Answer 8

Do the b^2 - 4ac next

Answer 9

so 169-88 ?

Answer 10

Yes which = 81

-13 +- sqrt(81) all over 2

Answer 11

Now... is the sqrt (81) an integer "a perfect number"

Answer 12

so x equals 13 and 9 ?

Answer 13

No I will help you

Answer 14

thank you so much

Answer 15

sqrt(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

Answer 16

i get it now... omg thank you so much!

Answer 17

NP

Answer 18

Be 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

Answer 19

i have like 3 more but im going to try them on my own first and if i get stuck ill post it (: thanks!

Answer 20

ok so like you said i theres more steps for ones that are not "perfectly" well i got -7+- sqrt -33/ 2

Answer 21

i know -33 is not a perfect square so what would i do next?

Answer 22

factor out any "perfect" squares from 33.....there aint none..

so leave it as is

Answer 23

thanks!

Answer 24

if 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

Answer 25

okay thank you (: so i need help with another problem please? It says to solve by factoring or using quadratic equation; the problem given is (x-6) (5x+9) = 0 how would i do this?

Answer 26

ok... solve by factoring OR quad formula right?

Answer 27

yes :)

Answer 28

since the equation is ALREADY in a factored form; we use the rule that anything times zero = 0.

that means that when (x-6) = 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?

Answer 29

x= 6 and 9/5ths ?

Answer 30

your "numerals" are correct, but check the "signs" for me.

Answer 31

x= 6 x= -9/5ths ?

Answer 32

exactly, that was very good :)

Answer 33

thanks 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

Answer 34

thanks for your help though!

Answer 35

youre quite welcome; and I hope you find better days ahead.

Answer 36

That problem you were doing with -33, can you post the actual problem?

Answer 37

x2+7x-18=0

Answer 38

b^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

Answer 39

a = 1
b = 7
c = -18

Answer 40

wouldn't it be -72?

Answer 41

No... 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)

Answer 42

oooo okayy thank youu!

Answer 43

Let me know what you get for answers

Answer 44

49-72 = -23 right?

Answer 45

or do i add 49 and 72 ?

Answer 46

7^2 - 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

Answer 47

so then its going to be -7+ 11/2 and then -7 -11/2??

Answer 48

Yes.. do the (-7 + 11) first then /2
and (-7 - 11) then /2

Answer 49

answers are 2 and -9 ?

Answer 50

Yes... Do you have another one?

Answer 51

thank you! yes i have one more a^2-3a = 6 = 0

Answer 52

OK, 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.

Answer 53

so is it going to be -3 +- sqrt 9-24/ 2 ? 9-24 is -15 which isn't perfect so what would i do next ?

Answer 54

(before I check that... you had an equal sign before the 6, is it a plus or a minus

Answer 55

plus sorry

Answer 56

OK,
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

Answer 57

algebra 1 and yes a little bit

Answer 58

I am surprised that this is your next problem.. If you are sure you entered it correctly.

3 +- sqrt(9-24)/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

Answer 59

im lost with the "i" and "-"

Answer 60

Are you sure your problem was correct?

I doubt you have learned about imaginary numbers. that is why I think your problem may be incorrect.

Answer 61

no the problem is correct its an extra credit problem

Answer 62

I 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.

Answer 63

thank you so much mam!
god bless <3
i will be sure to add your email to contacts (:

Answer 64

You 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)