McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
7. Solving ax^2+bx+c=0
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Exercise 43 Page 514

Recall that a quadratic equation can have zero, one, or two solutions.

See solution.

Practice makes perfect

We begin by recalling that a quadratic equation can have zero, one, or two solutions. ax^2+bx+c = 0 ↙ ↓ ↘ Zero One Two If there are two solutions, we need to consider the context of the real-world situation to determine whether one or both solutions answer the given question.

Extra

Examples of Real-World Situations
Let's see some examples of real-world situations that use quadratic equations.

Example 1

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.

- t^2+3t+10 = 0
t^2-3t-10 = 0
â–¼
Solve for t
t^2-5t+2t-10 = 0
t(t-5)+2t-10 = 0
t(t-5)+2(t-5) = 0
(t-5)(t+2)=0
lct-5=0 & (I) t+2=0 & (II)
lt=5 t+2=0
lt=5 t=-2

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.

Example 2

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.

- p^2+10p+479 = 500
- p^2+10p-21 = 0
â–¼
Solve for p
p^2-10p+21 = 0
p^2-7p-3p+21 = 0
p(p-7)-3p+21 = 0
p(p-7)-3(p-7) = 0
(p-7)(p-3) = 0
lcp-7=0 & (I) p-3=0 & (II)
lp=7 p-3=0
lp=7 p=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.