Question 204169: I need help with a problem, could someone please help me. How do create a real life situation that fits into the equation (x+3)(x-5)=0. I know how to solve the equation, but I'm not sure how I can create a real life situation to fit the equation to express the situation as the same equation.
Found 2 solutions by stanbon, RAY100:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Graph the function f(x) = (x+3)(x-5) just to see what it looks like.
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Let "y" be "amount of money in the bank.
Or let "y" be the height of a diver above the water.
Or let "y" be altitude of an airplane above 500' in the air.
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Let "x" be time.
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The graph shows you that y decreases from -inf to x=1.
y is negative between x= -3 and x = 5;
y increases from x=1 to inf.
y is positive after x = 5
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Use that information to describe a realistic scenario modeled
by y = (x+3)(x-5)
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Cheers,
Stan H.
Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! Usually a quadratic equation deals with area, length * width
.
In this case we could ask for the area if a rectangle, that is similar to a square(x*x), but has length longer by 3, and width shorter by 5
However, this will result in a number, not zero
.
If we expand (x+3)(x-5) = x^2 -2x -15 =0,, and set it equal to y, we have a parabolic function, (y+15) =x^2 -2x,,or (y+15) +1 = x^2 -2x +1,,,,
(y+16) = (x-1)^2
.
We could then ask for the roots of the function.
.
In real life application, this could be temperature, with limits at y=0,,
or population trends, with critical points at y=0,
or the definition of projectile motion(football,or bullet), and y=0 would define range.
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