Problem 1

This is adapted from a problem by Jaireet Chahal.

Solve for all real values of x such that

|x – 1| – |x + 1| + x < 0

Problem 2

This is from the 2011 AMC 10B, Problem 19. Solve for real values of x such that:

√(5|x| + 8) = √(x² – 16)

As usual, watch the video for a solution.

Two problems about absolute values

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Answer To Absolute Value Inequality

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Problem 1

If x &geq; 1, then x – 1 and x + 1 will be greater than or equal to 0 so we can remove the absolute value signs.

x &geq; 1
|x – 1| – |x + 1| + x < 0
x – 1 – (x + 1) + x < 0
x < 2
So
1 &leq; x < 2

For -1 &geq; x < 1, then x – 1 is less than 0 so the absolute value sign makes it an opposite value, whereas x + 1 is greater than or equal to 0 so we can remove its absolute value sign.

-1 &geq; x < 1
|x – 1| – |x + 1| + x < 0
-(x – 1) – (x + 1) + x < 0
0 < x
So
0 < x < 1

Finally, if x < -1, both absolute value signs will force an opposite to the inside terms.

x < -1
|x – 1| – |x + 1| + x < 0
-(x – 1) – (-(x + 1)) + x < 0
So
x < -2

So we have 3 conditions:

1 &geq; x < 2
or
0 < x < 1
or
x < -2

The first two inequalities can be combined, so we have:

0 < x < 2
or
x < -2

Problem 2

√(5|x| + 8) = √(x² – 16)

Square both sides of the equation and note x² = |x|². Since we are squaring an equation it is possible we may end up with extraneous roots, so it is important to check possible solutions in the original equation. So let’s proceed.

5|x| + 8 = x² – 16
5|x| + 8 = |x|² – 16
0 = |x|² – 5|x| – 24
0 = (|x| – 8)(|x| + 3)
|x| = 8 or |x| = -3

But |x| &geq; 0 by definition, so the second solution must be excluded. Thus |x| = 8 gives x = -8 or 8. Both of these do work in the original equation so both sides are equal to √48 = 4√3. The product of the roots is:

-8 × 8 = -64

Reference

Problem 1 Jaireet Chahal
https://twitter.com/ChahalJaireet/status/1691638659542319359

Problem 2 AoPS 2011 AMC 10B, Problem 19
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19