SOLUTION: Sir/Madam, A pleasant day! Please help me understand how to solve this problem. Find the value of k so that the equation is: 2x^2+kx-15=0, has one root of 3 Thank you ve

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Sir/Madam, A pleasant day! Please help me understand how to solve this problem. Find the value of k so that the equation is: 2x^2+kx-15=0, has one root of 3 Thank you ve      Log On


   

Question 767354: Sir/Madam,
A pleasant day! Please help me understand how to solve this problem. Find the value of k so that the equation is:
2x^2+kx-15=0, has one root of 3
Thank you very much for your help!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+2x%5E2+%2B+k%2Ax+-+15+=+0+
I would use the quadratic formula:
+x+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+2+
+b+=+k+
+c+=+-15+
+x+=+%28-k+%2B-+sqrt%28+k%5E2+-+4%2A2%2A%28-15%29+%29%29+%2F+%282%2A2%29+
+x+=+%28-k+%2B-+sqrt%28+k%5E2+%2B+120+%29%29+%2F+4+
---------------------------------
If one of the roots is +3+, I can say:
+3+=+%28-k+%2B+sqrt%28+k%5E2+%2B+120+%29%29+%2F+4+
+-k+%2B+sqrt%28+k%5E2+%2B+120+%29+=+12+
+sqrt%28+k%5E2+%2B+120+%29+=+k+%2B+12+
Square both sides
+k%5E2+%2B+120+=+k%5E2+%2B+24k+%2B+144+
+24k+=+-24+
+k+=+-1+ answer
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I'll see if this works
+x+=+%28-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2+%2B+120+%29%29+%2F+4+
+x+=+%28+1+%2B-+sqrt%28+1+%2B+120+%29%29+%2F+4+
+x+=+%28+1+%2B+11+%29+%2F+4+
+x+=+12%2F4+
+x+=+3+
and, also.
+x+=+%28+1+-+sqrt%28+1+%2B+120+%29%29+%2F+4+
+x+=+%28+1+-+11+%29+%2F+4+
+x+=+-10%2F4+
+x+=+-5%2F2+
+2x+=+-5+
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+%28+x-3+%29%2A%28+2x+%2B+5+%29+=+0+
+2x%5E2+%2B+%28-1%29%2Ax+-+15+=+0+
OK