+29 Can anyone help me?
a) x^2-2x-3=0 ; x^2-2x-3> 0
b) x^2-2IxI-3=0 ; x^2-2IxI-3> 0
c) Ix^2-2x-3I=3
+128474 a) x^2-2x-3=0 ; x^2-2x-3> 0
Factoring x^2-2x-3=0 we have that (x -3)(x +1) =0....and setting each factor to 0, we have that x = 3 and x = -1
For the second part, the polynomial will be greater than 0 either between x= -1 and x = 3 or in the two intervals on either side of this one. If we let x = 0 in the polynomial, note that it makes its value = -3. And this is less than 0. And since 0 falls in the "middle" interval above, the two "outside" intervals will make the inequality true. Thus, x < -1 and x > 3 are the two intervals that "work."
b) x^2-2lxl-3=0 ; x^2-2lxl-3> 0
This one is a little trickier. Let's start by rearranging x^2-2lxl-3=0 ......add 2lxl to both sides....this gives us
x^2 - 3 = 2lxl divide both sides by 2....so we have
(x^2 - 3 )/ 2 = l x l
We have two equations here....either ... (x^2 - 3 )/ 2 = x or - (x^2 - 3 )/ 2 = x
Working with the first one, multiply both sides by 2...we have...
x^2 - 3 = 2x rearranging gives us
x^2 -2x - 3 = 0 and the solution to this is the same as in (a) above. Notice something, however.....x =3 is OK in the original equation, but x = -1 doesn't "work" In effect, we have an extraneous solution.
Now, for the second part, we have - (x^2 - 3 )/ 2 = x ...multiply both sides by -2...this gives us
x^2 - 3 = -2x add-2x to both sides
x^2 +2x - 3 = 0 factoring gives (x +3) (x - 1) = 0 so either x = -3 or x = 1. Note that x = 1 is also extraneous, but x = -3 is OK.
Here's a graph of the equation....https://www.desmos.com/calculator/7rptum7sk3
Notice that it = 0 only at x = -3 and x = 3. This graph will also help us solve the second part of (b). Notice that the function is > 0 on the intervals x< -3 and x > 3. The odd thing about this graph is that it almost looks like a parabola except for the "kink" at the "vertex." This is due to the absolute value in the function.
c) lx^2-2x-3l=3
To save some time, I'm just going to present the graphs of y = lx^2-2x-3l and y = 3. Their intersection points will give us a solution.
Here are the graphs......https://www.desmos.com/calculator/9gcr9w4tay
Notice that the intersection points occur at about x = -1.646, x = 0, x = 2 and x = 3.646. We could have proved this algebraically, but I'm running out of gas, at this point !!!
+128474 a) x^2-2x-3=0 ; x^2-2x-3> 0
Factoring x^2-2x-3=0 we have that (x -3)(x +1) =0....and setting each factor to 0, we have that x = 3 and x = -1
For the second part, the polynomial will be greater than 0 either between x= -1 and x = 3 or in the two intervals on either side of this one. If we let x = 0 in the polynomial, note that it makes its value = -3. And this is less than 0. And since 0 falls in the "middle" interval above, the two "outside" intervals will make the inequality true. Thus, x < -1 and x > 3 are the two intervals that "work."
b) x^2-2lxl-3=0 ; x^2-2lxl-3> 0
This one is a little trickier. Let's start by rearranging x^2-2lxl-3=0 ......add 2lxl to both sides....this gives us
x^2 - 3 = 2lxl divide both sides by 2....so we have
(x^2 - 3 )/ 2 = l x l
We have two equations here....either ... (x^2 - 3 )/ 2 = x or - (x^2 - 3 )/ 2 = x
Working with the first one, multiply both sides by 2...we have...
x^2 - 3 = 2x rearranging gives us
x^2 -2x - 3 = 0 and the solution to this is the same as in (a) above. Notice something, however.....x =3 is OK in the original equation, but x = -1 doesn't "work" In effect, we have an extraneous solution.
Now, for the second part, we have - (x^2 - 3 )/ 2 = x ...multiply both sides by -2...this gives us
x^2 - 3 = -2x add-2x to both sides
x^2 +2x - 3 = 0 factoring gives (x +3) (x - 1) = 0 so either x = -3 or x = 1. Note that x = 1 is also extraneous, but x = -3 is OK.
Here's a graph of the equation....https://www.desmos.com/calculator/7rptum7sk3
Notice that it = 0 only at x = -3 and x = 3. This graph will also help us solve the second part of (b). Notice that the function is > 0 on the intervals x< -3 and x > 3. The odd thing about this graph is that it almost looks like a parabola except for the "kink" at the "vertex." This is due to the absolute value in the function.
c) lx^2-2x-3l=3
To save some time, I'm just going to present the graphs of y = lx^2-2x-3l and y = 3. Their intersection points will give us a solution.
Here are the graphs......https://www.desmos.com/calculator/9gcr9w4tay
Notice that the intersection points occur at about x = -1.646, x = 0, x = 2 and x = 3.646. We could have proved this algebraically, but I'm running out of gas, at this point !!!
