document.write( "Question 175234: A Student claims that the equations sq -x =3 has no solution since the square root of a negative number does not exist. Why is this argument wrong? \n" ); document.write( "

Algebra.Com's Answer #130309 by gonzo(654) $\"\"$ $\"About$

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the student is wrong.
\n" ); document.write( "here's why.
\n" ); document.write( " $\"sqrt%28-x%29+=+3\"$
\n" ); document.write( "is the original equation.
\n" ); document.write( "---
\n" ); document.write( "if you square both sides of the equation you get:
\n" ); document.write( " $\"%28sqrt%28-x%29%29%5E2+=+3%5E2\"$ which becomes:
\n" ); document.write( "-x = 9 which becomes x = -9 once you multiply both sides of the equation by (-1).
\n" ); document.write( "---
\n" ); document.write( "to prove the answer is correct you substitute the value of -9 for x in the original equation.
\n" ); document.write( " $\"sqrt%28-%28-9%29%29+=+3\"$
\n" ); document.write( "which becomes:
\n" ); document.write( " $\"sqrt%289%29+=+3\"$
\n" ); document.write( "which becomes:
\n" ); document.write( "3 = 3 proving the value of x = -9 is good.
\n" ); document.write( "---
\n" ); document.write( "i believe the student was reacting to the fact that there was a negative sign under the square root and you can't take the square root of a negative number.
\n" ); document.write( "---
\n" ); document.write( "if x is negative, however, than a minus of a minus is a plus making the solution valid.
\n" ); document.write( "---
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