Quadratic equation and absolute value .

[siddscool19]

Quadratic equation and absolute value...

Consider equation |x²-2x-3|=m, m belongs to R
If the given equation has no solution then find the interval in which the m lies.



2. Homework Equations -Given in question



3.
Since this equation is in modulus there would be two equations:
One is : x²-2x-3=m
Second : x²-2x-3= -m

Discrimant=b²-4ac

Discriminant of 1st equation= 16+4m
Discriminant of 2nd equation= 16-4m

Discrimant for both the equations should be less than 0 (so that the roots are imaginary and there is no solution)

After solving D<0
I get m<-4
and m>4

I would have to take intersection of these equations to get the final answer. But there intersection would be a null set.

But the answer is not null set :(

Please tell me all my mistakes in my solution. Because there are similar problems in which also i am facing problems but since it wouldn't be good to ask all problems I am asking one only. :)

 

[Mark44]

You were spot on almost to the end. Here is where you fell down.

I would have to take intersection of these equations

No, you want the union of those inequalities. The solutions to your quadratic equations will be complex if D < 0, which happens if m < - 4 OR m > 4.

 

[Hurkyl]

Since this equation is in modulus there would be two equations:
One is : x^2-2x-3=m
Second : x^2-2x-3= -m

I think this is where you made your mistake.
You wrote down two equations, but you did not provide any annotation to go with them; how exactly do these equations relate to the original one?

You relation you assumed is

"x=a is a solution to |x^2 - 2x - 3| = m" if and only if both "x=a is a solution to x^2 - 2x - 3 = m" and "x=a is a solution to x^2 - 2x - 3 = -m"​

which I'm sure you can see is not true.


With care, you could "think" your way to the proper relation, but it's probably easier to appeal to the definition of absolute value and do the algebraic manipulations.

(The definition of absolute value contains two cases -- so this would mean that you have to break your problem up into the corresponding two cases, and solve each case separately)

 

[siddscool19]

Can you help me with the first few steps ?
I think that would help me learn better since I am new to this topic..

(Also if you know any useful resources related to absolute value and quadratic equation please let me know It would be a lot of help )

Between I didn't get why you think that I assumed this
"x=a is a solution to |x^2 - 2x - 3| = m" if and only if both "x=a is a solution to x^2 - 2x - 3 = m" and "x=a is a solution to x^2 - 2x - 3 = -m"
?

What I did is used |m|=m and |-m| is also equal to m.

So x^2-2x-3=m
and x^2-2x-3= -m

 

[HallsofIvy]

Can you help me with the first few steps ?
I think that would help me learn better since I am new to this topic..

(Also if you know any useful resources related to absolute value and quadratic equation please let me know It would be a lot of help )

Between I didn't get why you think that I assumed this
"x=a is a solution to |x^2 - 2x - 3| = m" if and only if both "x=a is a solution to x^2 - 2x - 3 = m" and "x=a is a solution to x^2 - 2x - 3 = -m"
?

As Hurkyl and siddscool19 have already told you, this is your mistake. |x|= 2 if x= 2 or -2, not "and".
""x=a is a solution to |x^2 - 2x - 3| = m" if either "x=a is a solution to x^2 - 2x - 3 = m" or "x=a is a solution to x^2 - 2x - 3 = -m"

What I did is used |m|=m and |-m| is also equal to m.

So x^2-2x-3=m
and x^2-2x-3= -m

That is wrong. x^2-2x-3= m or x^2- 2x- 3= -m.

 

[Hurkyl]

As Hurkyl and siddscool19 have already told you, this is your mistake. |x|= 2 if x= 2 or -2, not "and".
""x=a is a solution to |x^2 - 2x - 3| = m" if either "x=a is a solution to x^2 - 2x - 3 = m" or "x=a is a solution to x^2 - 2x - 3 = -m"

Only if. You need to add another condition to make this true if you use 'if'.

 

[Hurkyl]

Can you help me with the first few steps ?

Just remember the definition of absolute value breaks into two cases:

|z| = \begin{cases} z &amp; z \geq 0 \\ -z &amp; z &lt; 0 \end{cases}

Therefore, when |x^2 - 2x - 3| is involved, the first step is often

"x^2 - 2x - 3 \geq 0" or "x^2 - 2x - 3 &lt; 0"

and the second step is usually

Assume x^2 - 2x - 3 \geq 0