document.write( "Question 737342: I'm very confused on the substitution method, so I will write the question as it appears on the assignment.
\n" ); document.write( "Solve the following equation using the substitution method.
\n" ); document.write( "\"x%5E4-5x%5E2-14=0\"
\n" ); document.write( "Complete the following steps to solve the above equation:
\n" ); document.write( " a. Let u=x^2, substitute the variables, and write the new equation.
\n" ); document.write( " b. Now factor the new equation, apply the zero-product principle, & solve for u.
\n" ); document.write( " c. Now substitute u=x^2 to solve for x.
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Algebra.Com's Answer #450308 by Susan-math(40)\"\" \"About 
You can put this solution on YOUR website!
Since\"u=x%5E2\" then \"u%5E2=%28x%5E2%29%5E2=x%5E4\"
\n" ); document.write( "So \"x%5E4+-5x%5E2+-14=0\" is the same as \"u%5E2+-+5u+-14+=+0\". Which easily factors to be
\n" ); document.write( "\"%28u+-+7%29%28u+%2B+2%29+=+0\"
\n" ); document.write( "Set each factor to zero to solve.
\n" ); document.write( "\"u-7=0++or+u%2B2+=0\"
\n" ); document.write( "This gives us u = 7 or u = -2. \r
\n" ); document.write( "\n" ); document.write( "But we don't want u, we want x. Time to substitute using\"u=x%5E2\"
\n" ); document.write( "\"x%5E2=7+or+x%5E2=-2\"\r
\n" ); document.write( "\n" ); document.write( "So we have \"x=sqrt%287%29\"\"x=sqrt%287%29\",\"x=i%2Asqrt%282%29\",\"x=-i%2Asqrt%282%29\"
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