SOLUTION: Twice the square of an integer is 3 less than 7 times the integer. Find the integer. 2x^ = 1st integer 7x-3 = 2nd integer 2x^=7x-3 2x^ + 7x + 3 = 0 (2x+1)(x+3) 2x+1 =

Question 434850: Twice the square of an integer is 3 less than 7 times the integer. Find the integer.
2x^ = 1st integer
7x-3 = 2nd integer
2x^=7x-3
2x^ + 7x + 3 = 0
(2x+1)(x+3)
2x+1 = 0 x+3=0
2x=-1 x=-3

x=-1/2 & x = -3
this is as far as i could get? Is this right? We are to solve this using the GSSC model.

Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Twice the square of an integer is 3 less than 7 times the integer. Find the integer.
2x^2 = 1st integer
7x-3 = 2nd integer
2x^2=7x-3
2x^2 + 7x + 3 = 0 **********
(2x-1)*(x-3) = 0
x = +1/2, 3 Only your signs are wrong.
x^ is meaningless, the exponent has to be entered. It isn't always 2.
---------------
(2x+1)(x+3)
2x+1 = 0 x+3=0
2x=-1 x=-3

x=-1/2 & x = -3
this is as far as i could get? Is this right? We are to solve this using the GSSC model.
I don't know what the GSSC model means.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Twice the square of an integer is
3 less than 7 times the integer.
Find the integer.
------------
Equations:
2x^2 = 7x-3
---
2x^2-7x+3 = 0
Factor:
----
2x^2-6x-x+3 = 0
2x(x-3)-(x-3) = 0
----
(x-3)(2x-1) = 0
x = 3 or x = 1/2
=========================
Cheers,
Stan H.
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