SOLUTION: Please help me with this! 12x^2-5x-7=0, using completing the square. Thank you!

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Please help me with this! 12x^2-5x-7=0, using completing the square. Thank you!      Log On


   

Question 904912: Please help me with this!
12x^2-5x-7=0, using completing the square.
Thank you!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+12x%5E2+-+5x+-+7+=+0+
+12x%5E2+-+5x+=+7+
+x%5E2+-+%285%2F12%29%2Ax+=+7%2F12+
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Take 1/2 the coefficient of the x term,
square it, and add it to both sides
+x%5E2+-+%285%2F12%29%2Ax+%2B+%285%2F24%29%5E2+=+7%2F12+%2B+%285%2F24%29%5E2+
+x%5E2+-+%285%2F12%29%2Ax+%2B+25%2F576+=+336%2F576+%2B+25%2F576+
+x%5E2+-+%285%2F12%29%2Ax+%2B+25%2F576+=+361%2F576+
+%28+x+-+5%2F24+%29%5E2+=+%28+19%2F24+%29%5E2+
Take the square root of both sides
+x+-+5%2F24+=+19%2F24+
+x+=+24%2F24+
+x+=+1+
and
+x+=+5%2F24+=+-19%2F24+
+x+=+-14%2F24+
+x+=+-7%2F12+
----------------
check:
+%28+x+-+1+%29%2A+%28+x+%2B+7%2F12+%29+
+x%5E2+-+x+%2B+%287%2F12%29%2Ax+-+7%2F12+=+0+
+x%5E2+-+%285%2F12%29%2Ax+-+7%2F12+=+0+
+12x%5E2+-+5x+-+7+=+0+
OK