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Question 1148732: My math teacher didn't teach us how in the world to do this I am completely stuck.
The sum of 6 times a positive number and 1 is the same as the square of 1 less than the number.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! -----------------------------------------------------------------------
My math teacher didn't teach us how in the world to do this I am completely stuck.
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The sum of 6 times a positive number and 1 is the same as the square of 1 less than the number.
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p, the unknown positive number in the description
 ---------the description, literally turned into an equation.
Simplify and solve the quadratic equation.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Let the unknown number be "x".
Then 6 times this number is 6x; 6 times this number and 1 is (6x +1).
From the other side, "the square of 1 less than the number" is (x-1)^2.
These expressions, (6x+1) and (s-1)^2, "are the same", which means that they are equal at this x
6x+1 = (x-1)^2. (1)
The equation is done; now your task is to solve it.
6x+1 = x^2 - 2x + 1
x^2 - 8x = 0
x*(x-8) = 0.
It has two roots: x= 0 and x= 8.
Only x= 8 is POSITIVE. (x= 0 IS NOT positive (!) )
So your answer is x= 8.
CHECK. Left side of equation (1) is 6x+1 = 6*8+1 = 49.
Right side of equation (1) is (x-1)^2 = (8-1)^2 = 7^2 = 49.
Both sides are equal --- hence, the equation solved correctly.
The wording formulation is satisfied --- hence, the problem is solved CORRECTLY. (!)
Solved, explained, answered and completed.
Come again to this forum soon to learn something new (!)
Or to learn something better (!)
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