document.write( "Question 1183695: Find the value of K for which y=x+k is a tangent to the curve y=x²+5x+2 \n" ); document.write( "
Algebra.Com's Answer #814139 by robertb(5830)\"\" \"About 
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Without using calculus, a simpler solution would be as follows:\r
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\n" ); document.write( "\n" ); document.write( "Equate \"y+=+x%5E2+%2B+5x+%2B+2\" and \"y+=+x+%2B+k\".\r
\n" ); document.write( "\n" ); document.write( "===> \"x%5E2+%2B+5x+%2B+2+=+x+%2B+k\", since at their intersection point(s) they're supposed to have the same y-values.\r
\n" ); document.write( "\n" ); document.write( "<===> \"x%5E2+%2B+4x+%2B+%282-k%29+=+0\".\r
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\n" ); document.write( "\n" ); document.write( "Since the line is supposed to be tangent to the curve, it means that there are two identical roots for the preceding equation. \r
\n" ); document.write( "\n" ); document.write( "This only happens when the discriminant is equal to 0, i.e.,\r
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\n" ); document.write( "\n" ); document.write( "\"4%5E2+-+4%2A1%2A%282-k%29+=+0\" ===> \"4%5E2+=+4%282-k%29\" ===> \"highlight%28k+=+-2%29\".\r
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\n" ); document.write( "\n" ); document.write( "Solved.
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