Quadratic functions anyone? h(x)=-x^2-6x-9 f(x)=2x^2+9x+4 g(x)=x^2+x-20
Do you just want those solved?
I can solve it but its gonna take a lil min
maybe lik 10 min
first those equations have to be a x and y not both x...are you sure those are the right equations
yes @lovingod786
@lovingod786
@wolf1728 Yeah..but I need to show a little work..
Ik im doing it rn
Ok so you're gonna have double arenthese like (x 3) (x 3)...all you have to do is figure out the signs like negative and positive to get the equation lik -x^2 -6x-9
nvm...ima just do it the long way
haha... I was confused for a second @lovingod786
so first youre gonna identify your A,B, and C in the equation....ax^2+bx+c=0
Then you're gonna plug those letters into the equation
would -x just be -1? @lovingod786
\[x=\frac{ -b+\sqrt{b^2 -4ac}}{ 2a }\]
No it wudnt...ur solving for x
so in the equation where it states -b..it would only b -b if your b from the original equation is positve.
but the b from the original is negative...
so therefore it wud b positve
But the B that is in the sqrt root wud b the original negative B
This is how your equation should look
\[x= \frac{ 6 + \sqrt{(-6^2) - 4 (-1)(-9)} }{2(-1)}\]
get it?
YES!!(:
You can finish it from here?
I will...try?
Oh and btw where the + is b4 he sqaure root...how I was taught was..its also a minus sign under the addition sign so you will come out with two answer and either or.
Lemme see you work ou tthe problem
The either or is basically the same thing but one you get the subtraction you solve for the addition with the same stuff
\[\frac{ a }{ b } + \frac{ a }{ b }\]
\[\frac{ a }{ b } - \frac{ a }{ b }\]
Thats basically the either or format
Do you have the answer??
no.. I have no idea what to do now @lovingod786
I thought I could but..
What happened
Where did you get stuck
I just don't know what to do... I don't know where to start and what to start..
Ok just look...ima gonna do step by step.
ok! I will write this down so I can look later on too!
\[x = \frac{ -b + \sqrt{b^2 - 4ac} }{ 2a }\]
Wait...do you understand where the A,B, and C.come fromwith the original equation?
yes!
ok i will continue now
\[x= \frac{ 6 + \sqrt{-6^2 - 4(9)} }{ -2 }\]
\[x = \frac{ 6+ \sqrt{-6^2 - 36} }{ -2 }\]
\[x = \frac{ 6 + \sqrt{36-36} }{ -2 }\]
\[x = \frac{ 6 + 0 }{ -2 }\]
\[x = \frac{ 6 }{ -2 } = -3\]
THANK YOU! That helped a lot!
When I do the other problems (tomorrow, it's late), would I be able to check my work with you? @lovingod786
Yeah just tag me like you're doin now
I will get a notification
Will do! So is that the final answer, -3?
Yeah
Yay! That was so much easier than I expected... You were very good at explaining! thank you!
You're welcome
does x=15.75 for b? @lovingod786
lol oh snap I didnt even plug inhold on
hah! it's ok!(:
\[h(x) = -x^2 -6x -9\]
\[h(-3)=-(-3)^2 -6(-3) - 9\]
\[h(-3) = -(9) + 18 + 9 \]
Answer should b 0
Dang it... so now what?
I think I am going to go to sleep and try again in the morning!! for b: x=15.75 for c: I didn't understand @lovingod786 Good night!
thank you anyways!
I got the wrong answer hold on
You should follow PEDMAS
So it should be..
\[h(-3) = (3)^2 +18 -9\]
then so exponents so the equation wud b 9+9 = 18
I even checked with mathway
18 is the final answer for the equation h(18)
h(x)=-x^2-6x-9 testing the -3 answer h(x) = -(-3^2) -6*-3 -9 h(x) = -9 +18 -9 h(x) =0 so -3 is a valid answer
lovingod786 you used the quadratic formula for equation (h) x = [6 +-sq root (36 -36)] / -2 x = 6 +- sq root(0) / -2 -3 is one valid root what is the other root?
What do you mean the other root? @wolf1728
Because since it is a zero you dont need two answers
Solving the equation with the quadratic formula: x = 6 +- sq root(0) / -2 Notice there is a "+-" before the sq root.
Yes, so both answers are -3 and -3
There's no two answers for this...because all you are doing is subtracting and adding the zero so you could just leave it as -3 but if you want to write it like how you doing it then go right ahead
The only reason I am stating both answers is that teachers always insist upon the student giving BOTH answers (even if the answers are equal).
My bad...You don't plug intothe equation as I see...The answer you got where the formula contains a square root is the answer
It seems you were testing the -3 answer with the equation h(−3)=(3)2+18−9 and then you say 18 is the final answer for the equation h(18) Actually, you miscalculated in setting up h(−3)=(3)2+18−9 it should be h(−3)=-(-3)2 -(6*-3) −9 h(−3)= -9 +18 -9 h(-3) = 0
Yes, plugging the number into the equation is just the way to test if the answer is correct
I got that answer at first but then I plugged it into mathway to double check and they gave me 18 for the whole equation
Well at least we have the answer for the first equation :-)
Lol need help with the rest??
I think we should stop here and maybe have sophadof do a little work too.
ok
ok see ya
I am working on it!!!(: sorry for not responding, i was at work @wolf1728 @lovingod786
Here's the second quadratic equation f(x)=2x^2+9x+4 with a, b and c put into the quadratic formula: x= [-9 +-sq root(81 -4*2*4)] / 2*2 Can you solve it from here?
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