Exponents+Absolute Value Problem !

chaitanya.bhansali wrote:If r^3 + |r| = 0, what all are the possible values of r??

Could someone correct me on the steps shown below when solving this problem?

1.) Balance Non absolute and absolute terms on both the sides of the equation:

|r| = -r^3

2.) now 2 cases arise:

case a) When r is +ve: r = -r^3
=>r+r^3 = 0
=>r(1+r^2) = 0
=>r=0 or (1+r^2 = 0)
hence, r=0 and other value is not possible.

case b) When r is -ve: -r = -r^3
=> r^3 -r=0
=> r(r^2-1) = 0
=> r=0 or r=-1 and r=1
Hence r=0, which is common in both the cases.

I know when we put r = -1 and r =0 in the equation, they both satisfy it!

What is wrong with the step-wise solution above?

Thanks.

Given : If r^3 + |r| = 0

Since |r| > 0 therefore r^3 < 0 for their Sum to be Zero

Which leaves you with only one case i.e. r must be lesser than or equal to zero

i.e. r^3 = r
i.e. r^3 - r = 0
i.e. r(r^2 - 1) = 0
i.e. either r = 0 or r^2 = 1
i.e. either r = 0 or r = -1 (because r must be -ve therefore we can't take r = +or-1)

chaitanya.bhansali wrote:If r^3 + |r| = 0, what all are the possible values of r??

Could someone correct me on the steps shown below when solving this problem?

1.) Balance Non absolute and absolute terms on both the sides of the equation:

|r| = -r^3

2.) now 2 cases arise:

case a) When r is +ve: r = -r^3

hence, r=0 and other value is not possible.

case b) When r is -ve: -r = -r^3

Hence r=0, which is common in both the cases.

I know when we put r = -1 and r =0 in the equation, they both satisfy it!

What is wrong with the step-wise solution above?

Thanks.

Another input to notice is Since you have taken two completely INDEPENDENT cases therefore you Don't have to identify the COMMON SOLUTIONS

Another thing to Notice is : When you are taking r as positive or negative then "r=0" should ideally not be part of the solution however I am assuming that you have considered r as NON-POSITIVE and NON-NEGATIVE in two mentioned cases.

chaitanya.bhansali wrote:If r^3 + |r| = 0, what all are the possible values of r??

Could someone correct me on the steps shown below when solving this problem?

1.) Balance Non absolute and absolute terms on both the sides of the equation:

|r| = -r^3

2.) now 2 cases arise:

case a) When r is +ve: r = -r^3
=>r+r^3 = 0
=>r(1+r^2) = 0
=>r=0 or (1+r^2 = 0)
hence, r=0 and other value is not possible.

case b) When r is -ve: -r = -r^3
=> r^3 -r=0
=> r(r^2-1) = 0
=> r=0 or r=-1 and r=1
Hence r=0, which is common in both the cases.

I know when we put r = -1 and r =0 in the equation, they both satisfy it!

What is wrong with the step-wise solution above?

Thanks.

The portion in red is an unnecessary -- and thus incorrect -- constraint.
The solutions for r do not need to satisfy both cases.
In fact, when we "open" an equation with absolute value into two cases, normally there will NOT be a solution that satisfies both cases.

You derived the following:
|r| = -r³.
Since |r| cannot be negative, |r| ≥ 0.
Thus, for the equation above to be valid, the righthand side must also be greater than or equal to 0:
-r³ ≥ 0
r³ ≤ 0.
This is how the value of r is constrained.

Thus, only solutions for r such that r ≤ 0 are valid.
Of your three possible solutions -- r=-1, r=0, and r=1 -- two satisfy the constraint that r≤0:
r=-1 and r=0.

Generally, when an equation has absolute value on only one side, the easiest way to determine whether a solution is valid is to test it in the original equation.
Of your three possible solutions for r, only r=-1 and r=0 satisfy the original equation.
For more on this issue, check my posts here:
https://www.beatthegmat.com/simple-yet-a ... 74979.html

Please note that I do not recommend algebra for the problem above.
If we test a few easy cases -- r=0, r=±1, r=±2 -- we should see rather quickly that the only possible solutions for r³ + |r| = 0 are r=-1 and r=0.