# Thread: Absolute value equations and inequalities

1. ## Absolute value equations and inequalities

Help with these would be much appreciated. Thanks!

Solve the equation graphically.
|4x-3| = 5√(x+4)
(4x-3) = 5√(x+4)
4x= 5√(x+4) + 3
x = (5√(x+4) + 3)/4??

-4x+3=5√(x+4)
x=(5√(x+4) -3)/-4

Ok, assuming I did the math correctly (feel free to correct), I am not sure how to "Solve Graphically."

Solve the inequality, express in interval notation.
|x-8|>4
x=12,
-x=-4 so x=4?
x=12,4 did I do this correctly??

2. Originally Posted by Såxon

Solve the inequality, express in interval notation.
|x-8|>4
x=12,
-x=-4 so x=4?
x=12,4 did I do this correctly??

$\displaystyle |x-8|>4$

case 1 : $\displaystyle x-8>4$

$\displaystyle x>12$

case 2 : $\displaystyle x-8<-4$

$\displaystyle x<4$

Thus the solution would be $\displaystyle (12,\infty)U(-\infty,4)$

3. Originally Posted by Såxon
Help with these would be much appreciated. Thanks!

Solve the equation graphically.
|4x-3| = 5√(x+4)
(4x-3) = 5√(x+4)
4x= 5√(x+4) + 3
x = (5√(x+4) + 3)/4??

-4x+3=5√(x+4)
x=(5√(x+4) -3)/-4

Ok, assuming I did the math correctly (feel free to correct), I am not sure how to "Solve Graphically."
To "solve graphically", graph! y= |4x-3| consists of two straight lines. As long as 4x-3>0 or x> 4/3, the graph is the line y= 4x- 3, a line with slope 4 and x-intercept (0, 4/3). As long as x< 4/3, the graph is the line y= -(4x-3)= 3- 4x, a line with slope -4 and x-intercept (0, 4/3).

$\displaystyle y= 5\sqrt{x+4}$ is part of $\displaystyle y^2= 25(x+ 4)$ which is a parabola. Specifically, it is the part above the x-axis. You solve the equation by finding the x-coordinate of the point at which those intersect. Doing it on a graphing calculator makes it almost mindless.

Solve the inequality, express in interval notation.
|x-8|>4
x=12,
-x=-4 so x=4?
x=12,4 did I do this correctly??
math addict has already shown where you made your mistake on this: you are to solve the inequality not the equation. x= 12 and -4 (not 4)satisfy |x-8|= 4.