Quadratic inequality

**tonie** · April 8th 2011, 12:41 PM

x^2 +2x -3>0
I don't understand when to use the union symbol After using my three test points. i know the answer is (x-1)(x+3) so x = -3 X= 1

**Plato** · April 8th 2011, 12:47 PM

Originally Posted by tonie

x^2 +2x -3>0
I don't understand when to use the union symbol After using my three test points. i know the answer is (x-1)(x+3) so x = -3 X= 1

It is clear that $x=0$ is not a solution.
So $[-3,1]$ is not in the solution, the complement of that set: $(-\infty,-3)\cup(1,\infty)~.$

**tonie** · April 8th 2011, 01:56 PM

Yes but kneed to understand the rule you use in finding out how to use union and when not to

**tonie** · April 8th 2011, 01:59 PM

Don't understand the rule when you should use union and when not to use union for the problem above

**Plato** · April 8th 2011, 02:36 PM

Originally Posted by tonie

Don't understand the rule when you should use union and when not to use union for the problem above

You are asking for a hard and fast rule which does not exist.
In general, quadric inequalities can have the whole line as a solution set, such as $x^2+1\ge 0~.$

Or the solution set can be between to real numbers, as in $(x+2)(x-3)\le 0$ the set is $[-2,3]$ .

Out side two numbers: $(x-2)(x+3)\ge 0$ , solution $(-\infty ,-3]\cup [2,\infty)$ .

But that is only a rule of thumb.

**tonie** · April 8th 2011, 09:46 PM

I am sorry but I am not getting it. I know it must be something simple because of the way you are responding to my question but I need to know when the correct time is to use a union or simply (3,1)?

**mathfun** · April 8th 2011, 10:59 PM

Let $f\left( x \right) = a{x^2} + bx + c$

If D>0, then f(x) has two real roots. Between those the sign of f(x) is the opposite of the sign of $a$ and outside them it has the same sign
If D=0, f(x) has the same sign with a except the point of zero.
If D<0, f(x) has the same sign with a

**HallsofIvy** · April 9th 2011, 05:38 AM

Originally Posted by tonie

I am sorry but I am not getting it. I know it must be something simple because of the way you are responding to my question but I need to know when the correct time is to use a union or simply (3,1)?

Are you saying that you do not know what the union symbol means?

The fact that x= -3 and x= 1 make the two sides equal tells you that the inequality is one way or the other through out the three intervals, $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . Plato told you that the solution cannot be (1, 3) because x= 0 is in that interval and 0 does not satisfy the inequality. You really should, then, check one point in each of the other intervals. x= 2 is in the interval $(1, \infty)$ . $2^2+ 2(2)- 3= 4+ 4- 3= 5> 0$ so x= 2 and, so, every number in $(1, \infty)$ satisfies it. x= -4 is in $(-\infty, -3)$ and $(-4)^2+ 2(-4)- 3= 16- 8- 3= 5> 0$ so x= -4 and, therefore, every number in $(-\infty, -3)$ satisifes the inequality.

That is, any number in $(-\infty, -3)$ or in $(1, \infty)$ will satisfy the inequality. A point is in $A\cup B$ if and only if it is in A or in B, by definition.

**tonie** · April 11th 2011, 08:16 AM

You answered the question. Thank you, for understanding what I was trying to convey. I assumed I broke it clearly down to what I did and didn't understand, but I guess people are in a rush and only wish to explain the answer. Thank you so much for being patient and throughly explaining everything.

Originally Posted by HallsofIvy

Are you saying that you do not know what the union symbol means?

The fact that x= -3 and x= 1 make the two sides equal tells you that the inequality is one way or the other through out the three intervals, $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . Plato told you that the solution cannot be (1, 3) because x= 0 is in that interval and 0 does not satisfy the inequality. You really should, then, check one point in each of the other intervals. x= 2 is in the interval $(1, \infty)$ . $2^2+ 2(2)- 3= 4+ 4- 3= 5> 0$ so x= 2 and, so, every number in $(1, \infty)$ satisfies it. x= -4 is in $(-\infty, -3)$ and $(-4)^2+ 2(-4)- 3= 16- 8- 3= 5> 0$ so x= -4 and, therefore, every number in $(-\infty, -3)$ satisifes the inequality.

That is, any number in $(-\infty, -3)$ or in $(1, \infty)$ will satisfy the inequality. A point is in $A\cup B$ if and only if it is in A or in B, by definition.

**johnny** · April 11th 2011, 08:47 PM

y = x² + 2x - 3 has coefficient 1 > 0 for x², hence the parabola opens up.
x² + 2x - 3 = (x + 3)(x - 1) > 0
x = -3 or 1
-3 < 1
∴ x < -3 or x > 1

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