document.write( "Question 1094302: Solving by substitution with u
\n" ); document.write( " $\"+x%5E2-3x-sqrt%28x%5E2-3x%29=2+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"+u+=+sqrt%28x%5E2-3x%29+\"$
\n" ); document.write( " $\"+u%5E2+=+sqrt%28x%5E2-3x%29+=+x%5E2-3x+\"$
\n" ); document.write( "creates quadratic = $\"+u%5E2-u+=+2+\"$
\n" ); document.write( " $\"+u%5E2-u-2=0+\"$
\n" ); document.write( " $\"+%28u-2%29%28u%2B1%29+\"$
\n" ); document.write( "u = 2,-1
\n" ); document.write( "going back and using those two outputs in $\"+sqrt%28x%5E2-3x%29+\"$
\n" ); document.write( " $\"+sqrt%28%282%29%5E2-3%282%29%29+\"$ and $\"+sqrt%28%28-1%29%5E2-3%28-1%29%29+\"$
\n" ); document.write( " $\"+sqrt%284-6%29+\"$ and $\"+sqrt%281%2B3%29+\"$
\n" ); document.write( "Finally, I get $\"+sqrt%28-2%29+\"$ and $\"+sqrt%284%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? \n" ); document.write( "

Algebra.Com's Answer #853952 by MathTherapy(10801) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "Solving by substitution with u\r\n" );
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document.write( "creates quadratic = \r\n" );
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document.write( "u = 2,-1\r\n" );
document.write( "going back and using those two outputs in \r\n" );
document.write( " and \r\n" );
document.write( " and \r\n" );
document.write( "Finally, I get  and \r\n" );
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document.write( "But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? \r\n" );
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document.write( "u = 2,-1 <=== This is okay!\r\n" );
document.write( "\"going back and using those two outputs in      and  <=== Here's where I guess \r\n" );
document.write( "                                                                                                                you got confused, and substituted 2 and - 1 for x in \r\n" );
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document.write( "But, u = 2, as you mentioned above, NOT x = 2. And, because you'd substituted u for  earlier, at this juncture, you\r\n" );
document.write( "need to BACK-SUBSTITUTE the value of u to get:\r\n" );
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document.write( "                    ---- Squaring both sides\r\n" );
document.write( "                    \r\n" );
document.write( "     \r\n" );
document.write( "(x - 4)(x + 1) = 0\r\n" );
document.write( "  x - 4 = 0          OR        x + 1 = 0\r\n" );
document.write( "       x = 4           OR              x = - 1 \r\n" );
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document.write( "Now, you have 2 values for x that you can CHECK to ensure that they're VALID and NOT EXTRANEOUS. \r\n" );
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document.write( "Also, u = - 1, as you mentioned above. And, because you'd substituted u for  earlier, at this juncture, you need \r\n" );
document.write( "to BACK-SUBSTITUTE the value of u to get:  \r\n" );
document.write( "Seeing that the square root of ANY expression is positive (> 0), it's obvious that u = - 1 is an EXTRANEOUS value. As such,\r\n" );
document.write( "x = 4, or x = - 1 (see above).

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