SOLUTION: The equation 2x^2-5x=-12 is rewritten in the form of 2(x-p)^2+q=0. What is the value of q?

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Question 1065310: The equation 2x^2-5x=-12 is rewritten in the form of 2(x-p)^2+q=0. What is the value of q?
Found 3 solutions by josgarithmetic, josmiceli, Edwin McCravy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
12x%5E2-5x%2B12=0


The form containing p,
2%28x%5E2-2px%2Bp%5E2%29%2Bq=0

2x%5E2-4px%2B2p%5E2%2Bq=0

Comparing the corresponding coefficients,
system%285=4p%2C12=2p%5E2%2Bq%29
Can you continue here and solve this system and find the value for q?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+2%2A%28+x+-+p+%29%5E2+%2B+q+=+0+
+2%2A%28+x%5E2+-+2p%2Ax+%2B+p%5E2+%29+%2B+q+=+0+
+2x%5E2+-+4p%2Ax+%2B+2p%5E2+%2B+q+=+0+
-----------------------------------
The given equation is:
+2x%5E2+-+5x+=+-12+
+2x%5E2+-+5x+%2B+12+=+0+
------------------------------
+-4p+=+-5+
+p+=+%28-5%29%2F%28-4%29+
+p+=+5%2F4+
------------------
+2p%5E2+%2B+q+=+12+
+2%2A%28+5%2F4+%29%5E2+%2B+q+=+12+
+2%2A%28+25%2F16+%29+%2B+q+=+12+
+q+=+12+-+50%2F16+
+q+=+12+-+25%2F8+
+q+=+96%2F8+-+25%2F8+
+q+=+71%2F8+
-------------------------
check:
+2%2A%28+x+-+p+%29%5E2+%2B+q+=+0+
+2%2A%28+x+-+5%2F4+%29%5E2+%2B+71%2F8+=+0+
+2%2A%28+x%5E2+-+%2810%2F4%29%2Ax+%2B+25%2F16+%29+%2B+71%2F8+=+0+
+2x%5E2+-+%2810%2F2%29%2Ax+%2B+25%2F8+%2B+71%2F8+=+0+
+2x%5E2+-+5x+%2B+96%2F8+=+0+
+2x%5E2+-+5x+%2B+12+=+0+
+2x%5E2+-+5x+=+-12+
OK

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
The equation 2x^2-5x=-12 is rewritten in the form of 2(x-p)^2+q=0. 
+2x%5E2-5x=-12
+2x%5E2-5x%2B12=0

So we have the identity

2%28x-p%29%5E2%2Bq=2x%5E2-5x%2B12

Substitute x = 0

2p%5E2%2Bq=12

q=12-2p%5E2

Substitute x=p

2%28p-p%29%5E2%2Bq=2p%5E2-5p%2B12

q=2p%5E2-5p%2B12

Equate the two expressions for q

12-2p%5E2 ≡ 2p%5E2-5p%2B12

-4p%5E2%2B5p ≡ 0

p%28-4p%2B5%29 ≡ 0

p=0 or -4p%2B5=0
             p=5%2F4

But p cannot be 0, since

2%28x-0%29%5E2%2Bq ≡ 2x%5E2-5x%2B12

2x%5E2%2Bq ≡ 2x%5E2-5x%2B12 cannot be an identity, since 
there is no x term on the left

q=12-2p%5E2
q=12-2%285%2F4%29%5E2
q=12-2%2825%2F16%29
q=12-25%2F8
q=71%2F8

So (p,q) = %28matrix%281%2C3%2C5%2F4%2C%22%2C%22%2C71%2F8%29%29

But the first

Edwin