Question

solve using subtitution y=-x^2-2x+8 y=x^2-8x-12

Answer 1

set them equal and solve

Answer 2

\[-x^2-2x+8 =x^2-8x-12\]

Answer 3

ok then what

Answer 4

then i got them on one side and got -2x^2-6x+20

Answer 5

hello?

Answer 6

hen factor and solve.

Answer 7

thats where i get stuck

Answer 8

Then*

Answer 9

are they -10 and 4

Answer 10

Do you know how to factor ?

Answer 11

because it has sum of -6 right

Answer 12

Yes ,you're right

Answer 13

ok then what

Answer 14

i plugged it in and got -2x^2+10x+4x+20

Answer 15

is this right

Answer 16

-2x^2 -10x + 4x + 20

Answer 17

then what?

Answer 18

Then make pairs.
2 left terms and 2 right terms

Answer 19

\[-x^2-2x+8 =x^2-8x-12\] don't start with a negative leading coefficient, that makes your life harder
\[2x^2-6x-20=0\] then divide by 2
\[x^2-3x-10=0\] factor\[ (x-5)(x+2)=0\]

Answer 20

ok im kind of lost

Answer 21

A GPS can help you to get back on track. ;)

Answer 22

factoring: \[x^2-3x-10\]1) find multiples of -10 who's sum in -3. Which is -5 and 2, then rewrite equation. \[x^2+2x-5x-10\]2) Do some grouping, and simpler factoring\[(x^2+2x)+(-5x-10)=x(x+2)-5(x+2)\]3) Notice anything in common..? We can now rewrite in the simplest factored form\[(x-5)(x+2)\]