document.write( "Question 135617: Write a quadratic equation for which the following are solutions for x: ± 3i
\n" ); document.write( "This one has me a bit confused. Here's what I get
\n" ); document.write( "x = ±i sqrt 9
\n" ); document.write( "x = ± sqrt -9
\n" ); document.write( "x^2=-9
\n" ); document.write( "x^2+9=0
\n" ); document.write( "That's not a complete quadratic equaiton, but I don't know what to do next or if I'm doing it right to begin with.
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Algebra.Com's Answer #99355 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
You sort of took a convoluted path to get there, but you did arrive at the right answer.\r
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\n" ); document.write( "\n" ); document.write( "Here's how I would have done it:\r
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\n" ); document.write( "\n" ); document.write( "A polynomial equation has a root \"x=a\" if and only if \"x-a\" is a factor of the polynomial. Furthermore, a polynomial equation of degree n has exactly n roots.\r
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\n" ); document.write( "\n" ); document.write( "You are given two roots, so you know that you must have a quadratic (degree 2) polynomial equation.\r
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\n" ); document.write( "\n" ); document.write( "The roots are \"3i\" and \"-3i\", therefore, the factors of the polynomial must be \"x-3i\" and \"x-%28-3i%29=x%2B3i\".\r
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\n" ); document.write( "\n" ); document.write( "Multiply the two factors together: \"%28x-3i%29%28x%2B3i%29=x%5E2%2B9\" (factorization of the difference of two squares in reverse)\r
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\n" ); document.write( "\n" ); document.write( "So your equation is \"x%5E2%2B9=0\".\r
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\n" ); document.write( "\n" ); document.write( "And, by the way, it most certainly is a 'complete' quadratic equation of the form \"ax%5E2%2Bbx%2Bc=0\": \"a=1\", \"b=0\", and \"c=9\". The only restriction on the possible values for the coefficients in \"ax%5E2%2Bbx%2Bc=0\" is that \"a%3C%3E0\". b and c can be any values, including zero.
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