Question

College algebra

Answer 1

\(\large x^2-6x >4x-x^2 \)

add \(x^2\) both sides

Answer 2

the reason I reposted was because my answer is not matching what was on the other post.

Answer 3

its fine

Answer 4

2x^2 -6x greater than 4x

Answer 5

\(\large x^2-6x >4x-x^2 \)

add \(x^2\) both sides

\(\large 2x^2-6x > 4x \)

Answer 6

subtract 4x both sides, so that we can see a clean quadratic

Answer 7

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

Answer 8

\(\large x^2-6x >4x-x^2 \)

add \(x^2\) both sides

\(\large 2x^2-6x > 4x \)

subtract \(4x\) both sides

\(\large 2x^2-10x > 0 \)

Answer 9

divide 2 thru out

Answer 10

I did not know you could do that because it was 10x.

Answer 11

\(\large x^2-6x >4x-x^2 \)

add \(x^2\) both sides

\(\large 2x^2-6x > 4x \)

subtract \(4x\) both sides

\(\large 2x^2-10x > 0 \)

divide 2 thru out

\(\large x^2-5x > 0 \)

Answer 12

ok

Answer 13

\(\large x^2-6x >4x-x^2 \)

add \(x^2\) both sides

\(\large 2x^2-6x > 4x \)

subtract \(4x\) both sides

\(\large 2x^2-10x > 0 \)

divide 2 thru out

\(\large x^2-5x > 0 \)

factor \(x\)

\(\large x(x-5) > 0 \)

Answer 14

ab > 0 means
1) a > 0, b >0
2) a < 0, b <0

Answer 15

right ?

Answer 16

Yes Sir

Answer 17

lol ok im no sir but ok

Answer 18

*blush* sorry.

Answer 19

so, we got x(x-5) > 0

case I :-
x > 0 , x-5 > 0

case II :-
x < 0, x-5 < 0

Answer 20

work both cases

Answer 21

Done and done

Answer 22

Thanks so much

Answer 23

perfect !

Answer 24

Nice.

Answer 25

case I :-
x > 0 , x-5 > 0
x >0, x > 5
x > 5 **************** SOLUTION


case II :-
x < 0, x-5 < 0
x < 0, x < 5
x < 0 ********* solution

Answer 26

those two slutions give us :-
x > 5 : (5, infty)
x < 0 : (-infty, 0)


when u combine, we get watever the anwer u showed

Answer 27

@ganeshie8 , one question to your last 2 posts.
On case 1, why did you pick the 5 as the solution and and on case 2 pick 0? Why could you not have don the reverse?

Answer 28

Sorry for all the ?'s but I just noticed it.

Answer 29

Very good question !
ping me when you're on, we'll discuss :)

Answer 30

@ganeshie8 , I got your message.

Answer 31

okie glad to hear :) so u did get why

x >0, x > 5

becomes

x > 5

Answer 32

Well, it would not hurt to go ahead and show me.

Answer 33

Wait, yea I understand that.

Answer 34

lol nothing to show as such,
just we need to pick the thing that SATISFIES both

Answer 35

\[x-5>0\]
x is greater than 5

Answer 36

x >0, AND x > 5

x > 0 wont satisfy x > 5,
but x >5 will satisfy both x>0 and x > 5

thats why we picked x > 5

Answer 37

So let me ask this. A way to determine it could be to plug in a number greater than AND less than zero. Whichever satisfies the original equation \[x(x-5)>0\] is the one to pick.

Answer 38

So, negative 1 would satisfy that. Thus, x has to be less than zero.

Answer 39

As for checking if
x is greater than 5
works, plug in 6 to see if it satisfies the original equation
x(x−5)>0

So, 6 is greater than zero, it works.

Answer 40

all negative numbers will satisfy, cuz they make both x and (x-5) negative
product of two negatives is positive

Answer 41

Right, but that shows that \[x <0\] would in fact work.

Answer 42

Right?

Answer 43

thats right !
now look for the other half

Answer 44

when, x >=0, wat wil work ?

Answer 45

Pluging in 4 to \[x(x-5)>0\] would give -4. That wont work!
Pluging in 5 to the same equation would give 0, so all # 5 and above!

Answer 46

nope

Answer 47

i mean, you're right, but not exactly correct

Answer 48

plugging in 5 wont work
you need to plug a number GREATER than 5

Answer 49

My bad, your right!

Answer 50

just making sure u see it lol, but i see u nailed the thing :)

Answer 51

You don't know how much I appreciate your great help!

Answer 52

np :D

Answer 53

you're wlcme :)