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Recall that a quadratic equation can have zero, one, or two solutions.
See solution.
We begin by recalling that a quadratic equation can have zero, one, or two solutions.
Let h(t)=- t^2+3t+10 be a function that gives the height of a ball t seconds after being thrown from a height of 10 meters above the ground. To find the time at which this ball hits the ground, we must equate the function to 0. h(t) = - t^2+3t+10 = 0 Next, let's solve the equation above for t.
As we can see, we have found two solutions, t=-2 and t=5. However, since t represents time, it cannot be negative and we discard t=-2. In consequence, we conclude that the ball will hit the ground 5 seconds after being thrown.
The function R(p)=- p^2+10p+479 gives the revenue of a company when they set the price of their product as p dollars. Let's find the price per product for which the company makes a profit of $500. R(p)=- p^2+10p+479 = 500 Let's solve the equation above by factoring.
LHS-500=RHS-500
LHS * (-1)=RHS* (-1)
Rewrite -10p as -7p-3p
Factor out p
Factor out -3
Factor out (p-7)
Use the Zero Product Property
(I): LHS+7=RHS+7
(II): LHS+3=RHS+3
From the above we conclude that the company makes a profit of $500 when they set the price of their product equal to either $3 or $7.