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Do you know what a,b and c are?
my guess is that a = x^2 b= 13x and c= 22 ?
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?
OOPS "in front"
yes i do -b plus or minus square root radical b squared minus 4ac all over 2a ?
Good.. so you put those in (I like to read it the opposite of b), it helps when you put in values.
-13 +- sqrt(13^2 - 4(1)(22)) all over 2
Do the b^2 - 4ac next
so 169-88 ?
Yes which = 81 -13 +- sqrt(81) all over 2
Now... is the sqrt (81) an integer "a perfect number"
so x equals 13 and 9 ?
No I will help you
thank you so much
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
i get it now... omg thank you so much!
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
i have like 3 more but im going to try them on my own first and if i get stuck ill post it (: thanks!
ok so like you said i theres more steps for ones that are not "perfectly" well i got -7+- sqrt -33/ 2
i know -33 is not a perfect square so what would i do next?
factor out any "perfect" squares from 33.....there aint none.. so leave it as is
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
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?
ok... solve by factoring OR quad formula right?
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?
x= 6 and 9/5ths ?
your "numerals" are correct, but check the "signs" for me.
x= 6 x= -9/5ths ?
exactly, that was very good :)
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
thanks for your help though!
youre quite welcome; and I hope you find better days ahead.
That problem you were doing with -33, can you post the actual problem?
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
a = 1 b = 7 c = -18
wouldn't it be -72?
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)
oooo okayy thank youu!
Let me know what you get for answers
49-72 = -23 right?
or do i add 49 and 72 ?
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
so then its going to be -7+ 11/2 and then -7 -11/2??
Yes.. do the (-7 + 11) first then /2 and (-7 - 11) then /2
answers are 2 and -9 ?
Yes... Do you have another one?
thank you! yes i have one more a^2-3a = 6 = 0
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.
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 ?
(before I check that... you had an equal sign before the 6, is it a plus or a minus
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
algebra 1 and yes a little bit
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
im lost with the "i" and "-"
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.
no the problem is correct its an extra credit problem
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 firstname.lastname@example.org 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.
thank you so much mam! god bless <3 i will be sure to add your email to contacts (:
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)