Last problem! Help for medal.
I need help with just one more inequality equation...
x^2 - x - 15 > 15

- anonymous

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- anonymous

x<-5 or x>6

- anonymous

Would you like me to explain?

- anonymous

Yes please

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## More answers

- anonymous

sure

- UsukiDoll

@_greatmath7 thanks for providing an explanation in addition to the direct answer. :)

- anonymous

Let's solve your inequality step-by-step.
x^2-x-15>15
Let's find the critical points of the inequality.
(Subtract 15 from both sides): x^2-x-15-15=15-15
You then receive:
x^2-x-30=0
(Factor left side of equation):(x+5)(x-6)=0
(Set factors equal to 0):x+5=0 and x-=0

- anonymous

thnx @UsukiDoll

- anonymous

x-6=0

- anonymous

Oh I see. Thank you very much! :)

- UsukiDoll

o_0
\[\LARGE x^2 - x - 15 > 15\]
first we subtract 15 from both sides
\[\LARGE x^2 - x - 15-15 > 15-15\]
\[\LARGE x^2 - x - 30 > 0\]
then we have to factor. our last term is -30 and the middle term is -1
6 x -5 is the only combination that can work for this quadratic equation because
6 x -5=-30
6-5 = 1
Therefore
\[\LARGE (x-6)(x+5) > 0\]
now we solve
\[\LARGE (x-6)(x+5) > 0\]
by splitting this up into two cases
\[\LARGE (x-6) > 0\]
\[\LARGE (x+5) > 0\]
solving for an inequality is the same as solving for an equation
\[\LARGE x-6 > 0\]
\[\LARGE x-6+6 > 0+6\]
\[\LARGE x > 6\]
\[\LARGE x+5 > 0\]
\[\LARGE x+5-5 > 0-5\]
\[\LARGE x > -5\]
time to switch the inequality sign
\[\LARGE x < -5\]

- UsukiDoll

on the number line since we have < we have an open circle
|dw:1439510309024:dw|

- UsukiDoll

for example x < -5
the mouth of < is going on the right, but actually we have to shade the left side on the number line according to the tail
similarly x > 6
the > is going left, but actually we have to shade to the right on the number line according to the tail.

- anonymous

I'm sorry, but I've gotten a bit lost on the sign switching step. Could you explain that again?

- UsukiDoll

I got a bit confused on that one too... I think it has something to do with the negative number.
I know that if you divide both sides by a negative number the inequality signs switch, so I'm assuming that since we have a negative number as one of the solutions, the inequality signs switch.
But what I've read is that only swapping sides, and multiplication/division by negative numbers switch the inequality signs
so x > -5
swapping sides
-5 < x
but it's a standard to have that variable x on the left hand side.

- UsukiDoll

ah..maybe switching x> -5 by swapping sides might be it?
-5 < x
but nobody writes it like that even though it's still the correct answer.

- anonymous

Oops duh. Thank you for your help as well. :)

- anonymous

So it would be 6 < x > -5

- UsukiDoll

x.x
I would just leave it as x< -5 and x >6

- anonymous

What confuses me is how
x > -5
turns into
x< -5

- UsukiDoll

it probably because x > -5 doesn't make sense. x can't be greater than -5
the number lines either look like |dw:1439512229503:dw| or |dw:1439512254356:dw|

- UsukiDoll

^ that's just an example by the way
the first number line's notation is
-2 < x < 3
in order from smallest to greatest
the second number line's notation is
x < -2 or x >3

- anonymous

If x is greater than 6, wouldn't it also be greater than -5?
(I'm sorry if I'm becoming cumbersome. I am just having a hard time understanding)

- UsukiDoll

no.
x > 6
x < -5
http://www.wolframalpha.com/input/?i=solve+x%5E2+-+x+-+15+>+15
I already mentioned that when putting the inequality on the number line it should either be
in the format of
-23
|dw:1439514398312:dw|

- UsukiDoll

you know what's off?
only multiplying or dividing with negative numbers switches the signs.. a part of me wants x> -5 to be that answer.
but stupid wolfram has it at
x <-5
if I switch the sides on x >-5 and change the inequality sign
-5

- UsukiDoll

another example
Example: 12 < x + 5
If we subtract 5 from both sides, we get:
12 - 5 < x + 5 - 5
7 < x
That is a solution!
But it is normal to put "x" on the left hand side ...
... so let us flip sides (and the inequality sign!):
x > 7
Do you see how the inequality sign still "points at" the smaller value (7) ?
And that is our solution: x > 7
source: http://www.mathsisfun.com/algebra/inequality-solving.html

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