Question

Find the solutions to the system.
y=x^2+8x+2
y=7x+4

(-2,-10) and (1,11)
(2,10) and (-1,-11)
(-2,-34) and (1,11)
or no solutions
Please help ive tried everything!! thank you

Answer 1

have you tried setting two two equations equal and solving for x first?

Answer 2

that is you need to first solve:
7x+4=x^2+8x+2

Answer 3

so far i have 7x=x^2+8x-2

Answer 4

you are one step to having it in the form
ax^2+bx+c=0

then you can use quadratic formula or easier to factored the side in which you have ax^2+bx+c for your problem

Answer 5

do you know how to subtract 7x on both sides?

Answer 6

x^2+(8x-7x)-2=0 (is what you have after subtracting 7x on both sides
I left the left hand expression for you to simplify

Answer 7

im sorry im working slow i will simplify it hold on a moment please

Answer 8

wouldnt it just be x^3-2=0?

Answer 9

well 8-7 isn't 0

Answer 10

i added the x that was left to the other x^2 and made it x^3

Answer 11

oh I didn't really you but a cube on that x :(

Answer 12

8x-7x is simply 1x
not x^3-x^2 or whatever you did

Answer 13

im sorry i was confused because i thought that putting 1x was wrong because its just the same as x and i thought the x^2 was a like term to x so i added them

Answer 14

1x is the same as x
x^2+x doesn't simplify to x^3 (we cannot add x^2 and x because they are not like terms)

Answer 15

im sorry i was confused, what is the next step?

Answer 16

well you know you have x^2+x-2=0 right?

Answer 17

do you know how to factored x^2+x-2?
or do you prefer to use the quadratic formula to solve x^2+x-2=0?

Answer 18

i dont know either of them very well but i know the quadratic formula probably better

Answer 19

\[ax^2+bx+c=0 \text{ implies you have solutions } x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 20

you need to identify a,b,c as a first step in your equation 1x^2+1x+(-2)=0

Answer 21

1=a, 1=b,-2=c ?

Answer 22

that is right now you want to plug those numbers into the formula I have above
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(1)\pm \sqrt{(1)^2-4(1)(-2)}}{2(1)}\]

Answer 23

now simplify that

Answer 24

like I would start with the operations inside the square root

Answer 25

1^2 is the same as 1
can you evaluate 1-4(1)(-2) ?

Answer 26

is it 6?

Answer 27

1-4(1)(-2)
1-4(-2)
1+8
9
I hope you can see that the inside of the square root doesn't simplify to 6 but it simplifies to 9


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\[x=\frac{-1 \pm \sqrt{1-4(1)(-2)}}{2(1)} \\ x=\frac{-1 \pm \sqrt{1+8}}{2(1)} \\ x=\frac{-1 \pm \sqrt{9}}{2(1)} \]
now that you evaluated the operations inside the square root
we can look at the actual square root part
what is the square root of the number inside (that is what is the square root of 9?)?

Answer 28

3?

Answer 29

yes because 3*3 is 9
so you have gotten the square root part out the way
so you should have something that looks like this:
\[x=\frac{-1 \pm 3}{2}\]

Answer 30

well and I already did the bottom multiplication for you that is
I rewrote 2(1) as 2 since 2 times 1 is 2

Answer 31

can you finish simplifying your x values from here?

Answer 32

im just realy lost to how were eventually going to find the answer to the question

Answer 33

well you have to finish simplifying your x coordinates of your solutions

Answer 34

oh okay i might need a second hold on please

Answer 35

(-2,1) ?

Answer 36

do you mean you have x=-2 or x=1 ?

Answer 37

yes

Answer 38

now find the corresponding y values for each
the simplest equation to use is the y=7x+4

Answer 39

for x=-2 you have y=7(-2)+4=?
for x=1 you have y=7(1)+4=?

Answer 40

(-2,-10) (1,11) ?

Answer 41

yep

Answer 42

thank you so so much!! :)

Answer 43

np

Answer 44

some problems take a little patience especially if you don't know where you are going
y=x^2+8x+2
y=7x+4
we wanted to know when these y's were the same
x^2+8x+2=7x+4 ( this is why we set up this equality and why we had to solve it )
then after you find the x's
you can find the y's by pluggin into either equation

Answer 45

that was just a summary of what we did

Answer 46

anyways goodnight