Use the quadratic formula to find the zeros of the function. y = 16x2 + 40x + 25
A. {-1.25, 1.25}
B. {-1.25}
C. {-4, 5}
D. {-0.8}

- anonymous

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- anonymous

Quadratic formula \[(-b \pm \sqrt{b^2-4ac})/2a\]
where a, b, c
\[ax^2+bx+c=0\]

- anonymous

Can you see why you need to use the quadratic formula now?

- anonymous

ok Go on the steps to solve it confuse me cuz there so long.

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## More answers

- anonymous

so in your case, 16=a, 40=b, 25=c. Plug them into the equation that Romero showed you.

- anonymous

You want to find the zeros of function Y right?
that means y=0
first do that
you get \[16x^2+40x+25=0\]

- anonymous

mkay wow go on that makes so much more sense than anything my teachers are saying does

- anonymous

like jessie said a =16 b=40 c=25
now plug them in the equation I gave you on top.

- anonymous

ok then i will then what do i do u dont have to give me the answer i just need the simple non confusing steps ur giving me.... ^^

- anonymous

The \[\pm\]means that you have to do one with + and one with - so it makes sense that you should get two values. You should be able to plug them in and whatever you get is the answer for X

- anonymous

ok ill plug them in for you but you work it out

- anonymous

ok i will i dont need the answer! lol just the steps!

- anonymous

a= 16 b=40 c=25
\[x= (-b \pm \sqrt{b^2-4ac})/2a\]
so if I plug them in I get
\[x= (-40 + \sqrt{40^2-4*16*25})/(2*16)\]
\[x= (-40 - \sqrt{40^2-4*16*25}) / (2*16) \]

- anonymous

so just do all the multiplication division etc and you should get your answers

- anonymous

OH!!!! Ok thank you thank you!! xD

- anonymous

I plugged the first numbers in and then i got 8.75 as my answer... but thats not on the options!

- anonymous

check your signs!!

- anonymous

i did!! thats why it took so long!

- anonymous

Ignore the rest, and look under the radical. Start 40^2, which is 1600. Then look at the rest of what is under the radical. Do -4x16x25. Do you get -1600? So the square root of 1600-1600 is 0.

- anonymous

Are we good to there?

- anonymous

yea i got that

- anonymous

but i didnt get the 1600 part for the first part i accidentally

- anonymous

put 4 because i squared the 40 then wasnt thinking and then i got the square root which was.. 40

- anonymous

so then would you do -40 + 32 ?

- anonymous

wait no im confused

- anonymous

Well remember, 2x16 is 32, but your not adding, you are dividing.

- anonymous

So You would have -40 plus the square root of all that junk that turned out to be zero.

- anonymous

-40/32 is -1.25

- anonymous

ohhhhh so u dont add?

- anonymous

wow ok thx i didnt even see that!!!

- anonymous

No. At the end, you set your equation OVER 2a. In other words, divide it by 2a.

- anonymous

ohhhhh ok?

- anonymous

http://www.regentsprep.org/Regents/math/algtrig/ATE3/quadsongs.htm
memorize the equation, if you're in Algerbra or Algerbra 2 you're gonna use it again!

- anonymous

algebra*

- anonymous

ok thx!!! ya im in Algebra IIthe only reason im bad is because of all the confusing steps!!!! :P but thx u helped me ALOT!!!

- anonymous

No problem.

- anonymous

What a pun

- anonymous

wait wait so i wrote the problem down like this:
x = (-b + -
Square root of…
B^2 – 4ac
/2a

- anonymous

and then were we doing b^2 - 4ac or somethin?

- anonymous

bc i didnt understand why i would divide by 2a again
which was 32

- anonymous

wait ok im confused UGH! i thought i got it!!!

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