Absolute Value Question: Solving |2x - 1| = x^2 with Step-by-Step Guide

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SUMMARY

The discussion focuses on solving the absolute value equation |2x - 1| = x^2. The correct approach involves breaking the equation into two cases: 2x - 1 = x^2 for x ≥ 1/2 and -(2x - 1) = x^2 for x < 1/2. The solutions obtained are x = 1 and x = -1, but only x = 1 is valid within the defined constraints. Graphical representation of the functions y = |2x - 1| and y = x^2 is recommended to visualize the points of intersection.

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Homework Statement


how do I solve this? I am confused...
|2x - 1| = x^2

Homework Equations





The Attempt at a Solution


when i tried it I ended up with a solution set of (1,-1). But the official answer is quite different so I am confused!
 
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The absolute value function is defined as follows:

|x| = x, x\geq 0

|x|= -x, x&lt; 0

But you have 2x-1 so replace the x in the absolute value for 2x-1. Try this and then solve both equations 2x-1=x^2 and -(2x-1)=x^2. Once you find your solutions, remember to be sure to scrap the solutions that aren't valid. That is, if you get an answer of -10 for 2x-1=x^2 then you know it's not valid because we are assuming 2x-1>0, or, x>1/2.
 
You could also approach this graphically.

Imagine the line y=2x-1, that's a line with a positive slope and intercepts the y-axis at (0,-1). Now, |2x-1| refers to the absolute values of y (i.e. |y|) so you have to reflect what is below the x-axis in the x-axis itself to get your |y|=|2x-1| which looks like a 'V' shaped graph. The function of the graph that you obtain can be defined as follows:

y = 2x-1 for x &gt;= \frac{1}{2}

and
y = 1 - 2x for x &lt;= \frac{1}{2}

Now all you have to do is sketch the graph of x^2 on top of that and simply determine where your V-shaped graph (y=|2x-1|) is equal to the graph of y=x^2 by finding the points of intersection.
 

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