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Quadratic inequalities

Sal solves x^2+3x>10. Created by Sal Khan.

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  • piceratops ultimate style avatar for user Liam Nickell
    Could you use the quadratic formula to solve a question like this?
    (14 votes)
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    • male robot johnny style avatar for user Thomas B
      You can use the quadratic equation to find the endpoints of the intervals that will be you solution, and would then need to test in which of those intervals the inequality is true. So in this case you could use it to find -5 and 2 [(-3 +- Sqrt(9+4(10)1))/2 = (-3 +- 7)/2 = -10/2 or 4/2]. This breaks up the number line into 3 intervals {x<-5, -5<x<2 and x>2} for a quadratic inequality, the solution will always either be the two outer intervals or the middle one (think about how a graph would look), so you can just check the middle. In this case you can check x =0, and since 0^2 + 3(0) is not greater than 10, you know it will be the outside intervals {x<-5 or x>2}
      (11 votes)
  • male robot hal style avatar for user Jr Mangcoy
    What would happen if the greater than turns into less than?
    (6 votes)
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    • piceratops ultimate style avatar for user T H
      x² + 3x < 10
      x² + 3x - 10 < 0
      (x + 5)(x - 2) < 0
      ( x + 5 > 0 and x - 2 < 0 ) OR ( x + 5 < 0 and x - 2 > 0 )
      ( x > -5 and x < 2 ) OR ( x < -5 and x > 2 )
      Because it is impossible that x < -5 and x > 2, our final answer is x > -5 and x < 2, which we can write as -5 < x < 2.
      (7 votes)
  • aqualine ultimate style avatar for user Hari Shankar
    how do we solve an inequation like x-3/x-5>0
    (1 vote)
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    • duskpin sapling style avatar for user Rohini
      When you have an inequality like this, you can't multiply the denominator with 0 and make the simplification easier. This is because you are unsure of whether the value of "x" is positive or negative. And due to this , if x was negative you'd have to flip the inequality but like I said you do not know for sure. So this how you do it :
      (x-3)(x+5) / (x-5)(x+5) > 0
      (x-3)(x+5) / (x^2 - 25) > 0
      Now you can multiply the denominator with zero since even if your "x" is negative is will become positive since it's been squared. So,
      (x-3)(x+5) > 0
      So , you'll have two values, x > 3 or x > -5 . Now the final part is figuring out the interval. When you have two values , the method I use to find the interval is using the wavy curve method ( It would be tough for me to explain it , better to understand it visually than by writing). According to the wavy curve method, when you have "x" is greater than ( >) something, then you leave the values between the two numbers and take all the other values.
      Since our values are -5 and 3, the solution will be -∞ .....-8,-7,-6,-5 and 3,4,5,6....+∞. When you write this in interval notation , you'll have (-∞,-5)U(3,+∞)
      (8 votes)
  • blobby green style avatar for user 😊
    how will you solve this x² + 8x +15 < 0
    (1 vote)
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    • starky ultimate style avatar for user MichelleW.
      Problem: x^2+8x+15<0

      First step: Factor out the inequality. (what times what equals 15 and when added together makes 8?)
      (x+3)(x+5)<0

      Step 2: Solve for x. This inequality has two answers.
      X can either be -3 or -5, since both, when plugged in for x, will make the inequality equal to zero.

      Step 3: Draw a number line with the points -3 and -5 plotted with hollow circles, since the inequality has a < sign.

      Step 4: Plug in -4 (representing the space in between the points -3 and -5 on the number line) into the (x+3)(x+5) and solve.
      (-4+3)(-4+5)= -1 The answer is negative.

      Step 5: Look at x^2+8x+15<0. The answer is suppose to be LESS than 0, which means that the answer has to be a NEGATIVE number. Since when -4 is plugged into the inequality, the answer is negative, the space in between the points -3 and -5 on the line is the SOLUTION.

      Step 6: Write the solutions to the inequality in the form of X<? or X>?. Since the answer is in between -3 and -5, the solutions should be X>-5 or X<-3.

      Final solutions: x>-5 or x<-3

      Hope this helped!!
      (6 votes)
  • starky sapling style avatar for user Paula Rózpide
    how would you solve this?:
    find the set of values of x for which
    3 - 5x - 2x^2 < 0
    (2 votes)
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  • leafers ultimate style avatar for user Aloizio Soares
    I thought that a quadratic inequality would generate an area like it happens in linear functions. is it possible?
    (1 vote)
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    • female robot ada style avatar for user Maryam Meetha
      That can be the case. If you have a quadratic inequality then you need to factorise it to work out your values of x. Once you have done this, you can plot them on a graph. If the inequality is > 0, than the area the inequality is representing will be be above the x -axis. However, if the inequality is < 0 than the area represented is below the x - axis.
      (3 votes)
  • blobby green style avatar for user Aditi.sadia.Rahman
    x^2>1 and x^2<1
    what are the values for x.
    (1 vote)
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  • blobby green style avatar for user dylanhoi13
    I'm new-ish to algebra and I have no idea how to approach this question: -2b^2+1000b-1000. Originally it's from these two equations which I merged: 2(a+b)=1000 and a+b+2ab<1500. I might've done a step wrong but I can't factor the first equation and when I apply the quadratic formula, it comes out with square roots.
    (1 vote)
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  • piceratops seed style avatar for user pushkardeore1501
    how will we solve (x-1)(x-2)(x-3) > 0
    (1 vote)
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  • blobby green style avatar for user megawatt2119
    What if I had a less than symbol
    (1 vote)
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Video transcript

Let's say that we want to solve the inequality x squared plus 3x is greater than 10. We want to figure out all of the x's that would satisfy this inequality. I encourage you to pause this video now. And I'll give you a hint. Try to manipulate the way that you would have if this was a quadratic equation. But then as you get to the end, try to reason through it, because the reasoning might departure a little bit from what you are used to. So I'm assuming you've given a go at it. So the first thing that we might want to do, just to get into a form that we're more comfortable with, is subtract 10 from both sides. If we subtract 10 from both sides, then on the left hand side, we're going to have x squared plus 3x minus 10 is still going to be greater than. If we add or subtract the same thing to both sides, it won't change the inequality. But it's now going to be greater than 0. 10 minus 10 is 0. Now, this gets us into a form that we're more used to seeing quadratic expressions in. If this was an equal sign right over here, we'd want to factor this thing. So let's just try to factor here too and see what happens. So we're going to factor it. We're going to think of two numbers whose product is negative 10 and whose sum is positive 3. And we've had a lot of practice doing this. If you think about the factors of 10, it's 1, 2, 5, and 10. 2 and 5 seem tempting, because their difference is 3. So if you have positive 5 and negative 2, that seems to work out. Positive 5 and negative 2. Their product is negative 10, their sum is positive 3. We could rewrite this as x plus 5. Let me do that in that yellow color so you see where this 5 is coming from. X plus 5 times x minus 2 is going to be greater than 0. Now, if this was an equality here, we would say well, how do we get this equals 0? If either of these things were equal to 0, then this entire expression would be equal to 0, because 0 times anything is 0. But we don't have an equality here. We have a greater than symbol. So let's think about how we could reason through this. And I'll do a little bit of an aside here. If I were to tell you that numbers a and b, and if I were take the product a times b, and if someone were to tell you that product is greater than 0, what do we know about a and b? Well, we know that they have to have the same sign. They're either both positive-- a positive times a positive is going to be a positive-- or they're both going to be negative-- a negative times a negative is a positive. It's going to be greater than 0. So we know the same thing here. Let me write it down. So we know either a is greater than 0 and b is greater than zero-- so either both of them are positive or both of them are negative-- or a is less than zero and b is less than zero. So we apply that same logic here. You could view this x plus 5 is a, you could view this x minus 2 as our b of the product of two things. The product is greater than 0. That means that either both of these expressions are positive or they're both negative. So let's write that down. I'll write it this way. So either both of these expressions are positive. So either x plus 5 is greater than 0 and x minus 2 is greater than 0-- let me write it this way-- or they're both negative. x plus 5 is less than 0 and x minus 2 is less than 0. So now let's think about all of these inequalities independently. But let's maintain this logic here of the and and the or. So let's look at this. Either they're both positive, so if both of these expressions are positive, what do we know about x? Well, if you subtract 5 from both sides of this inequality, you get x is greater than negative 5. And if you add 2 to this inequality, both sides of that inequality, you're going to get x is greater than 2. So if x is greater than negative 5 and x is greater than 2, what do we know about x? Well, any x that's greater than two is going to be greater than negative 5. So we could just simplify this right over here to say that x is greater than 2. So all of this, this is equivalent to saying x is greater than 2, because clearly anything that is greater than 2 will satisfy that. And both of these things have to be true. For example, x equal negative 4, it would satisfy this inequality but not this one. And so the and would break down. And negative 4 does not satisfy both of these. In order to satisfy both of these, you essentially have to satisfy that one. So this expression simplified to that. Now what about this? What about this statement right over here? Well, x plus 5 less than 0, subtract 5 from both sides. That is, x is less than negative 5. And add 2 to both sides of that inequality, you get x is less than 2. Now if x is less than negative 5 and x is less than two, what do we know about x? Well, that just means that x has to be less than negative 5. If it's less than negative 5, it's definitely going to be less than 2. And we got to remind ourselves that we have this or here. And that's essentially describing the solution set for this quadratic inequality here. x is going to be greater than 2 or x is going to be less than negative 5. And we could actually plot this solution set on a number line. So if this is our number line right over here, and let's say that this is 0. Let's say that's 1. 2 right over here. This is negative 1, negative 2, negative 3, negative 4, negative 5. So x could be greater than 2, not greater than or equal to, so I'm going to put an open circle here. So it could be greater than 2. Or it could be less than negative 5, not less than or equal to, so I'm going to put an open circle here. And so it could be less than that. So x could be any number. It could be negative 6, negative 6 would satisfy this. You could verify that. Negative 6 squared is 36 plus negative 18, which is going to be 18, which is greater than 10. Or you could have, say, a positive 3 would work. 3 squared is 9 plus another 9 is going to be 18, which once again, it is greater than 10.