document.write( "Question 649476: find k so that y=kx-4 and y=3x^2+x+2 intersect twice \n" ); document.write( "

Algebra.Com's Answer #407166 by htmentor(1343) $\"\"$ $\"About$

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find k so that y=kx-4 and y=3x^2+x+2 intersect twice
\n" ); document.write( "Setting the y-values equal, we have:
\n" ); document.write( "kx - 4 = 3x^2 + x + 2
\n" ); document.write( "Collect terms and set=0:
\n" ); document.write( "3x^2 + (1-k)x + 6 = 0
\n" ); document.write( "We can solve for x using the quadratic formula:
\n" ); document.write( "x = (-(1-k) +- sqrt((1-k)^2 - 72))/6
\n" ); document.write( "For there to be two solutions, the discriminant must be >0:
\n" ); document.write( "(1-k)^2 - 72 > 0
\n" ); document.write( "(1-k)^2 > 72
\n" ); document.write( "1-k > +/- sqrt(72)
\n" ); document.write( "Solving the inequality gives the two conditions on k:
\n" ); document.write( "k < 1 - sqrt(72)
\n" ); document.write( "k > 1 + sqrt(72)
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