document.write( "Question 1135573: For the function f(x) = |x2 - 3| - 2x, find the values of x such that f(x) = 0 \n" ); document.write( "

Algebra.Com's Answer #753219 by greenestamps(13209) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "I would solve the problem by rewriting the equation as

\n" ); document.write( " $\"abs%28x%5E2-3%29+=+2x\"$

\n" ); document.write( "Since an absolute value is always zero or positive, we can see from this form of the equation that any solutions will be positive.

\n" ); document.write( "Now, instead of looking at a graph of |x^2-3|-2x and looking for zeros, we graph |x^2-3| and 2x separately and look for intersections. Working this with pencil and paper, I would draw a quick sketch of the two functions. In this forum, we have a nice tool which draws accurate graphs:

\n" ); document.write( " $\"graph%28400%2C200%2C-5%2C5%2C-2%2C10%2Cabs%28x%5E2-3%29%2C2x%29\"$

\n" ); document.write( "The intersection with the lower x value occurs on the portion of the graph of |x^2-3| where the equation is y = -(x^2-3) = 3-x^2. So this solution is where

\n" ); document.write( " $\"3-x%5E2+=+2x\"$
\n" ); document.write( " $\"x%5E2%2B2x-3+=+0\"$
\n" ); document.write( " $\"%28x%2B3%29%28x-1%29+=+0\"$
\n" ); document.write( " $\"x+=+-3\"$ or $\"x+=+1\"$

\n" ); document.write( "We know our solutions will be positive; and the graph shows an intersection \"near\" x=1. So x=1 is one solution, which we can easily verify in the original equation.

\n" ); document.write( "The intersection with the larger x value is on the portion of the graph of |x^2-3| where the equation is y = x^2-3. So this solution is where

\n" ); document.write( " $\"x%5E2-3+=+2x\"$
\n" ); document.write( " $\"x%5E2-2x-3+=+0\"$
\n" ); document.write( " $\"%28x-3%29%28x%2B1%29+=+0\"$
\n" ); document.write( " $\"x+=+3\"$ or $\"x+=+-1\"$

\n" ); document.write( "Again we know our solutions have to be positive; and the graph shows an intersection near x=3. So x=3 is another solution, also easily verified in the original equation. \n" ); document.write( "

\n" );