SOLUTION: Hello, Can someone help me with this "Completing the Square" problem? I know there's a way of solving expressions that have a negative "x" squared, but I would really like some

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Hello, Can someone help me with this "Completing the Square" problem? I know there's a way of solving expressions that have a negative "x" squared, but I would really like some      Log On


   

Question 943223: Hello,
Can someone help me with this "Completing the Square" problem?
I know there's a way of solving expressions that have a negative "x" squared, but I would really like someone to explain the solution using the "(x-h)^2=p" form.
This is the problem, thank you in advance:
"Solve by completing the square: -4x^2+3x+7=0"
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I can show you the way I do it
+-4x%5E2+%2B+3x+%2B+7+=+0+
Subtract +7+ from both sides
+-4x%5E2+%2B+3x+=+-7+
Divide both sides by +-4+
+x%5E2+-%28+3%2F4+%29%2Ax+=+%28-7%29%2F%28-4%29+
+x%5E2+-%28+3%2F4+%29%2Ax+=+7%2F4+
Take +1%2F2+ of the coefficient of the
+x+ term, then square it, then add it
to both sides
-----------------------------------
+x%5E2+-%283%2F4%29%2Ax+%2B+%28-3%2F8%29%5E2+=+7%2F4+%2B+%28-3%2F8%29%5E2+
+x%5E2+-%283%2F4%29%2Ax+%2B+9%2F64+=+7%2F4+%2B+9%2F64+
+x%5E2+-%283%2F4%29%2Ax+%2B+9%2F64+=+112%2F64+%2B+9%2F64+
+x%5E2+-%283%2F4%29%2Ax+%2B+9%2F64+=+121%2F64+
Believe it or not, both sides are now perfect squares!
+%28+x+-+3%2F8+%29%5E2+=+%28+11%2F8+%29%5E2+
Take the square root of both sides
+x+-+3%2F8+=+11%2F8+
+x+=+14%2F8+
And, using the negative square root of the right side,
+x+-+3%2F8+=+-11%2F8+
+x+=+-1+
--------------
So, the factoring would be:
+%28+x+-+14%2F8+%29%2A%28+x+%2B+1+%29+=+0+
+x%5E2+-+%2814%2F8%29%2Ax+%2B+x+-+14%2F8+=+0+
+x%5E2+-+%28+6%2F8+%29%2Ax+-+14%2F8+=+0+
Multiply both sides by +-8+
+-8x%5E2+%2B+6x+%2B+14+=+0+
Divide both sides by +2+
+-4x%5E2+%2B+3x+%2B+7+=+0+
whew! This is a difficult one, but the
method is always the same