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cansomeone show me how to solve this? Use the quadratic formula to find the zeros of the function. y = 16x2 + 40x + 25 {-1.25, 1.25} {-1.25} {-4, 5} {-0.8

Mathematics
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do you recall the setup of the quadratic formula?
\[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
good, and the general form of a quadratic is: \[ax^2 + bx + c\] pull the numbers from the problem that relate to the abc parts and what do you get worked out?

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a=16 b=40 and 25=c
good, i notice that there are no - signs to worry about so that looks good so far
what do i do next?
plug in those numbers into their abc parts in the formula, the rest is just math for maths sake
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\] \[x=\frac{ -40 \pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\]
\[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] is that right?
looks good to me :)
ok then i get 40 pm 16-96/32?
but i would end up eith a negative number....
\[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{4^2(10)^2-4(4^2)(25)} }{ 2(16) }\] \[\frac{ -40\pm 4\sqrt{10^2-4(25)} }{ 2(16) }\] \[\frac{ -40\pm 4\sqrt{25(4)-4(25)} }{ 2(16) }\] \[\frac{ -40}{ 2(16) }\] \[\frac{ -40}{ 2(4.4) }\] \[\frac{ -10}{ 2(4) }\] \[\frac{ -5}{ 4 }\]
you lost me at the second one you did how did you get all the the ^2?
40^2 = (4*10)^2 = 4*10 * 4*10 = 4^2 * 10^2 but you dont have to do exactly what i did
uuuuh.....im sorry you lost me all i know is im suppose to square 40 is that what you were doin?
i was doing some manipulations so that i didnt have to worry about computing large numbers and such; i simplified it down to where i recognized the sqrt part goes to zero.
but 40^2 - 4(16)(25) = 0
ui havent learned taht i think im suppose to just use the big numbers
then math it out for maths sake, youll get big numbers that finally reduce and simplify
i tried but i keep getting teh wrong number when i have 40^2-4(16)(5) i get 0 right? i get 1,600 -1,600 and i cant square thaty
what does 1600 - 1600 equal?
0 -_- i cant square 0
sure you can, 0*0 = 0
oooooh......
so -40 pm 0/32?
\[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{1600-1600} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{0} }{ 2(16) }\] \[\frac{ -40\pm 0 }{ 2(16) }\] \[\frac{ -40}{ 2(16) }\]
yes /32 is fine as well
THANK YOU SO MUCH!!!!!
youre welcome
so its -40/32?
yes, which a calculator should be able to put into decimals for you
-1.25?
correct
thank you!

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