Question

Who is good with inequalities? Because I need help. I will give a medal to the person I believe answers my question I distributed first

Here is the question: x(7-x)>8

I distributed to get 7x-x>8

I don't know what to do after this

Answer 1

you didn't distribute...

\[x(7-x)>8 \Rightarrow 7x-x^{2}>8\]

Answer 2

That is what I meant

Answer 3

@pgpilot326

Answer 4

som next, get everything on one side and factor...

Answer 5

so

Answer 6

Okay then, this leads to a quadratic equation.

Answer 7

\[7x-x squared >8\]

Answer 8

There are going to be two solutions

Here are the choices 2,5,-1, and zero

Answer 9

move everything to the RHS

\[0>x^{2}-7x+8\]
now factor

the only way the inequality can be satisfied is if the RHS is negative

Answer 10

Right side?

Answer 11

oops. same idea but correct factoring.

Answer 12

How did u get the above answer?

Answer 13

They did the right side to change the \(-x^2\) into a positive. They could have also done it on the left side by turning the > into a < and multiplying everything by \(-1\).

Answer 14

sorry, it doesn't factor like that.
once factored, plot the zeros on a number line. you'll get 3 regions. see if any of those regions satisfy the inequality

Answer 15

you can always use the quadratic formula

Answer 16

Multiply everything by -1, the inequality sign switches directions, and then you solve the equation regularly?

Answer 17

yeah or just rewrite...

\[0>x^{2}-7x+8 \Rightarrow x^{2}-7x+8<0\]

Come on @e.mccormick

Answer 18

Just allows you to change the left and right and get it into a format you might be a little more comfortable with.

Answer 19

What should I do next?

Answer 20

Do you understand how you will use the quadratic in this problem? pgpilot326 is quite right in saying you will need it.

Answer 21

Make it into parenthesis (x+#) (x+#) Something like this

Answer 22

It would be x- x- a number for each

Answer 23

Yes.
So put \(x^{2}-7x+8=0\) into the quadratic like you normally would. Get the exact answers. Then put the exact answers in where the equation was on the original:
\(x^{2}-7x+8<0\implies (x\pm ?_1)(x\pm ?_2)<0 \)

Answer 24

I am thinking of x-2 x-5 and it isn't working because I remember that both of the numbers have to add up to -7x and when multiplied together equal positive 8

Answer 25

Which is why you need the quadratic. You will have a root.

Answer 26

\[\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 27

Do I do square root signs for both of them?

Answer 28

@pgpilot326 And yes, flipping it after getting it with the > works very well too! I was just pointing out the math of doing so.

@Comm.Dan Plug in the numbers and see.

Answer 29

K Thx for the reminder of the formula I forgot it because I haven't done quadratics in a long time

Answer 30

@pgpilot326

Answer 31

Thx for the help so far @pgpilot326 and @e.mccormick

Answer 32

@e.mccormick i understand but really there doesn't need to be any math involved.

saying that 0 > -1 is the same as saying that -1 < 0. there is math to rework but i think it's better to understand the simplicity first.

Answer 33

don't get me wrong, what you did is perfectly valid and I mean no disrespect @e.mccormick

Answer 34

\[7\pm \sqrt-39\] over 2

Answer 35

-39?

Answer 36

\((-7)^2-4(1)(8)\)

Answer 37

Sorry, my bad I looked and I made a mistake

Answer 38

Yah, forgot to square it first. =)

Answer 39

?
\[\frac{7 \pm \sqrt{(-7)^{2}-4(1)(8)}}{2(1)}=\frac{7 \pm \sqrt{49-32)}}{2}=\frac{7 \pm \sqrt{17)}}{2}\]

Answer 40

I got:
\(\dfrac{7+\sqrt{17}}{2},\dfrac{7-\sqrt{17}}{2}\)

You should too after fixing it.

Answer 41

I did, too. I got the same after realizing my mistake

Answer 42

OK, so:
\(\left(x-\dfrac{7+\sqrt{17}}{2}\right)\left(x-\dfrac{7-\sqrt{17}}{2}\right) > 0\)

Where is that true? Try a value above, between and below.