Question

Please help! Conic Solutions? Name and Calculate the solutions for each system.
16x^2 - 3y^2 = -11
8x - y = -11

Thank you so much!

Answer 1

Please help! Thank you!

Answer 2

@jdoe0001

Answer 3

welcome to openstudy!!

Answer 4

The answer is supposed to be: (-1,3) and (-2,-5) but i dont know how

Thanks!

Answer 5

np!
im leah btw!

Answer 6

Hi leah! im srini!

Answer 7

I fanned u so u can message me now...os works like facebook a little:)

Answer 8

Oh okay ! awesome!

Answer 9

im here to help if u ever need anything!!!

Answer 10

aww thank you! im still confused about this problem though like how they even got those answers :(

Answer 11

im not sure either :(
oh well...

Answer 12

are u online school too?

Answer 13

no.

Answer 14

oh...haha

Answer 15

Are you supposed to be solving the equations?

Answer 16

no she figured it out...lol

Answer 17

yes!

Answer 18

I still didnt understand it.

Answer 19

I think thats one part.

Answer 20

http://www.mathway.com/answer.aspx?p=calg?p=8xSMB15-SMB15ySMB15SMB01SMB15-11?p=1?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=SolveSMB15theSMB15EquationSMB15forSMB15x


You can go there for caculations

Answer 21

but i dont understand the step by step things.

Answer 22

have you done system of equations yet?

Answer 23

yeah. but i keep getting weird answers

Answer 24

ok... so, if you solve the bottom equation for "y"
8x - y = -11
y = 8x+11

Answer 25

okay.
then u plug it in right?

Answer 26

so we use that \(\bf y = 8x+11\) in the above equation, yes

Answer 27

but when i do that, i end up with a weird answer.

Answer 28

well, let's see
$$
16x^2-3(8x+1)^2=-11\\
\text{using the binomial theorem}\\
(8x+1)^2 = 64x^2+16x+1
$$

Answer 29

thus
\(\bf 16x^2-3(64x^2+16x+1)=-11\\
16x^2-192x^2-48x-3=-11\\
-176x^2-48x-3+11=0 \implies -176x^2-48x+8=0
\)

Answer 30

okay but how does that equal those coordinates?

Answer 31

thank you for helping me btw!

Answer 32

from there we'd need to use the quadratic formula to get the "x" values
keep in mind is a quadratic equation, degree is 2, so "x" will have 2 values

Answer 33

\(\bf \text{quadratic formula}\\
x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\)

Answer 34

wait.. lemme get common factors first

Answer 35

\(\bf -176x^2-48x+8=0 \implies -8(22x^2+6x-1) = 0\\ \implies 22x^2+6x-1=0\)

Answer 36

so we'll use the quadratic formula on that one

Answer 37

idk why but im not getting the correct answer.

Answer 38

$$\bf 22x^2+6x-1=0\\
x= \cfrac{ - 6 \pm \sqrt { 6^2 -4(22)(-1)}}{2(22)}\\
x= \cfrac{ - 6 \pm \sqrt { 36+88}}{44}\\
x= \cfrac{ - 6 \pm \sqrt { 124}}{44} \implies x= \cfrac{ - 6 \pm 2\sqrt {31}}{44}\\
$$

Answer 39

$$\bf x= \cfrac{ - \cancel{6} \pm \cancel{2}\sqrt {31}}{\cancel{44}}
\implies x= \cfrac{ - 3 \pm \sqrt {31}}{2}
$$

Answer 40

woops

Answer 41

$$\bf x= \cfrac{ - \cancel{6} \pm \cancel{2}\sqrt {31}}{\cancel{44}}
\implies x= \cfrac{ - 3 \pm \sqrt {31}}{22} $$

Answer 42

and those are 2 values for "x"

Answer 43

yeah but that doesnt match with the answers :(

Answer 44

what answers do yo have as choices?

Answer 45

I mean, that can be written in decimal form too

Answer 46

he answer is supposed to be: (-1,3) and (-2,-5

Answer 47

the one above is just the rational form

Answer 48

lemme check

Answer 49

how do i get it out of the rational form?

Answer 50

to plot it?

Answer 51

the graphs only meet at one point
even though the parabola is symmetric
I was thinking they'd meet at 2 points, the 2 "x" values, but they only meet at 1
near the parabola's vertex

Answer 52

and is not the coordinates you have there

Answer 53

to plot them, well
just solve for "y"
one equation will be
y = 8x + 1

the other equation will be
\(\bf y = \sqrt{\cfrac{16x^2+11}{3}} \)

Answer 54

since they meet at one point, means one of the "x" values won't work for it

Answer 55

anyhow, I just checked the decimal values, the positive root from the quadratic formula result is the good one

Answer 56

which is in decimal x = 0.12

Answer 57

and if you want the "y"
just use that in the y = 8x + 1 => y = 8(0.12) +1

Answer 58

the 2nd root of the quadratic equation is in decimal form -0.39
which yields the "y" coordinate of -2.12
so (-0.39, -2.12)
BUT
if you get the vertex of the parabola equation, you'll notice that point is below the vertex, so therefore it gets ruled out, and the positive root for 0.12 for "x" is the good one