Download Pre-Calculus: Unit 1 Lesson 6 Answer Key 1. Solve the equation 1 4

Pre-Calculus: Unit 1 Lesson 6 Answer Key
1. Solve the equation 3 x + 4 = x 2 − 1 graphically using the intersection method.
Identify the two equations that
are graphed.
10
9
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4
3
2
1
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
1
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5
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10
-2
-3
-4
-5
-6
-7
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-9
-10
x = -2.233 and 2.988
2. Solve the equation 3 x + 4 = x 2 − 1 graphically using the intercept method by
making one side of the equation equal to zero. Identify the equation that is
graphed.
2
Graph 3 x + 4 − x +1 = 0 and find the x-intercepts.
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4
3
2
1
-10
-9
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-2
-1
0
-1
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-8
-9
x = -2.233 and 2.988
-10
1
2
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9
10
3. How does finding the intersection points in problem 1 relate to finding the
intercepts in problem 2? Explain.
You will arrive at the same solutions, the intercept method takes one extra
step. You have to set your equation equal to zero before you can graph the
function and find the x-intercepts.
4. Consider the equation x 2 − 4 = c .
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1
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-9
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-1
0
-1
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-10
(a) Find a value for c which this equation has 4 solutions. Prove your
statement algebraically, numerically, or graphically.
One possible answer: c = 3
(b) Find a value for c that has 2 solutions. Prove your statement algebraically,
numerically, or graphically.
One possible answer: c = 5
(c) Find a value for c that has no solution. Explain.
One possible answer: c = -1
10
5. Solve the equation 2 x − 1 = 4 − x 2 algebraically.
2x −1 = −(4 − x 2 )
2x −1 = 4 − x 2
2x −1 = −4 + x 2
x 2 + 2x − 5 = 0
x = 1.499
x 2 − 2x − 3 = 0
(x − 3)(x +1)
x = −3.499(extraneous)
x = 3(extraneous)
x = −1
6. Solve the equation x(2 x + 5) = 4( x + 7) algebraically.
x(2x + 5) = 4(x + 7)
2x 2 + 5x = 4x + 28
2x 2 + x − 28 = 0
(2x − 7)(x + 4) = 0
x = 3.5
x = −4
7.
Norman window has the shape of a square with a semicircle mounted on it. Find
the width of the window if the total area of the square and semicircle is 200
square feet.
Area of square: x • x
Area of semi-circle:
π (0.5x)2
2
200 = x 2 + 0.5π (0.5x)2
x = 11.984
The width of the window is 11.984 feet.