anonymous
  • anonymous
Last problem! Help for medal. I need help with just one more inequality equation... x^2 - x - 15 > 15
Mathematics
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anonymous
  • anonymous
Last problem! Help for medal. I need help with just one more inequality equation... x^2 - x - 15 > 15
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x<-5 or x>6
anonymous
  • anonymous
Would you like me to explain?
anonymous
  • anonymous
Yes please

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anonymous
  • anonymous
sure
UsukiDoll
  • UsukiDoll
@_greatmath7 thanks for providing an explanation in addition to the direct answer. :)
anonymous
  • 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
  • anonymous
anonymous
  • anonymous
x-6=0
anonymous
  • anonymous
Oh I see. Thank you very much! :)
UsukiDoll
  • 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
  • UsukiDoll
on the number line since we have < we have an open circle |dw:1439510309024:dw|
UsukiDoll
  • 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
  • anonymous
I'm sorry, but I've gotten a bit lost on the sign switching step. Could you explain that again?
UsukiDoll
  • 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
  • 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
  • anonymous
Oops duh. Thank you for your help as well. :)
anonymous
  • anonymous
So it would be 6 < x > -5
UsukiDoll
  • UsukiDoll
x.x I would just leave it as x< -5 and x >6
anonymous
  • anonymous
What confuses me is how x > -5 turns into x< -5
UsukiDoll
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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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