MHB Proving Equivalence of f(x) and g(x)

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To determine the equivalence of the functions f(x) and g(x), both must be simplified. The function f(x) simplifies to 3x² + 5x - 7, while g(x) simplifies to 3x² + 3x - 7. Since the coefficients of the x² terms are the same, but the coefficients of the x terms differ, f(x) and g(x) are not equivalent. Therefore, the two functions do not yield the same results. The conclusion is that f(x) and g(x) are not equivalent functions.
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Consider the two functions f(x)=(x²+3x+10)+(2x²+2x-17) and g(x)=(4x²+4x+4)-(x²+x+11). If they are equivalent, prove they are; if they are not equivalent, prove they aren't.
 
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The first function can be simplified by adding like terms: [math]f(x) = 3x^2 + 5x - 7[/math]. Do the same for g(x). Do they come out the same?

-Dan
 
pappoelarry said:
Consider the two functions f(x)=(x²+3x+10)+(2x²+2x-17) and g(x)=(4x²+4x+4)-(x²+x+11). If they are equivalent, prove they are; if they are not equivalent, prove they aren't.
f(x)=(x²+3x+10)+(2x²+2x-17)= (x^2+ 2x^2)+ (3x+ 2x)+(10- 17)
Can you finish that ?

g(x)=(4x²+4x+4)-(x²+x+11)= (4x^2- x^2)+ (4x- x)+ (4- 11)
Can you finish that?

Are they the same?
 

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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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