SOLUTION: show that 2x^2+p=2(x-1) has no real roots if p>-3/2

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Question 1133428: show that 2x^2+p=2(x-1) has no real roots if p>-3/2
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The discriminant is +b%5E2+-+4a%2Ac+ when the equation is
+a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0+
Your equation:
+2x%5E2+-+2%2A%28+x-1+%29+%2B+p+=+0+
+2x%5E2+-+2x+%2B+2+%2B+p+=+0+
++a+=+2+
+b+=+-2+
+c+=+p+%2B+2+
--------------------
The discriminant is:
+4+-+4%2A2%2A%28+p+%2B+2+%29+
+4+-+8p+-+16+
+-8p+-+12+
If the discriminant is negative, there are
no real roots
+-8p+-+12+%3C+0+
+-8p+%3C+12+
+p+%3E+-12%2F8+ ( reverse the inequality sign when dividing by negative )
+p+%3E+-3%2F2+ for no real roots