Question

What are the x-intercepts of the following quadratics (solve only by factoring):
a) x2-10x+6y=-16
b) x2-10x+10y=0
c) x2-10x+12y=11
BEST ANSWER GETS MEDAL!!

Answer 1

Do you know how to factor?

Answer 2

Are you sure the y's are in these problems?

Answer 3

This is exactly the problem that was written, yeah.

Answer 4

I would know how to factor without the y's this is weird

Answer 5

Yeah haha

Answer 6

Wait an x intercept is when y=0

Answer 7

So all the y's would go away in these problems and than you would have to set all of them equal to 0 before you factor

Answer 8

Does this make sense?

Answer 9

I think so

Answer 10

Okay so on the first one once you set it equal to 0 you will get?

Answer 11

Plug 0 into y because every x intercept is where y=0 so you will have?

Answer 12

X^2-10x+6(0)=-16

Answer 13

\[x^2-10x+6y=-16\]

That's the first one?

Answer 14

So i got to the last one, and i'm left with x^2 - 10x = 0

Answer 15

\[x^2-10x+12y=11\]

You got:

\[x^2-10x = 0\]

Is there anyway to factor this?

(x )(x )

Answer 16

not sure

Answer 17

can one of them be 0?

Answer 18

I don't see a way in factoring it. I would start off like kayders said. Plug in 0 for y, because the x-intercept is where y=0

Answer 19

i did that,

Answer 20

it was originally x^2 - 10x + 10y = 0

Answer 21

Oh, this is for b.). My mistake, I thought it was for the last one

Answer 22

oh my bad sorry i was wrong

Answer 23

Then yes, you can use 0 as one of them. I see it working

Answer 24

What happened to the 11?

Answer 25

If you have an x in bothparts youcan take out an x(x-10) would be left

Answer 26

So x=10

Answer 27

And 0

Answer 28

The 11 was for c.)
This issue was for b.)

And yes, 11 and 0 are the intercepts

Answer 29

I mean 10 and 0