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## Algebra (all content)

### Unit 9: Lesson 12

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

## Want to join the conversation?

• Could you use the quadratic formula to solve a question like this? • 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}
• What would happen if the greater than turns into less than? • how do we solve an inequation like x-3/x-5>0
(1 vote) • 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,+∞)
• how will you solve this x² + 8x +15 < 0
(1 vote) • 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!!
• how would you solve this?:
find the set of values of x for which
3 - 5x - 2x^2 < 0 • I thought that a quadratic inequality would generate an area like it happens in linear functions. is it possible?
(1 vote) • x^2>1 and x^2<1
what are the values for x.
(1 vote) • 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) • how will we solve (x-1)(x-2)(x-3) > 0
(1 vote) • What if I had a less than symbol
(1 vote) 