1. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
Solve x2 + 2 = –3x using the quadratic formula.
x2 + 3x + 2 = 0
Add 3x to each side and write in standard form.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–3 ± (–3)2 – 4(1)(2)
2(1)
Substitute 1 for a, 3 for b, and 2 for c.
x =
–3 ± 1 2
Simplify.
x =
–3 + 1 2
–3 – 1 or
Write two solutions.
x = –1 or x = –2
Simplify.
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2. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
(continued)
Check: for x = –1 for x = –2
(–1)2 + 3(–1)
(–2)2 + 3(–2)
1 –
4 –
0 = 0
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3. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–4 ± – 4(3)(–8)
2(3)
Substitute 3 for a, 4 for b, and –8 for c.
–4 ± 6
x = or
Write two solutions. – 6
–4 –
Use a calculator. x – 6
–4 – or x
x –2.43
Round to the nearest hundredth.
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4. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second.
Step 1:  Use the vertical motion formula.
h = –16t2 + vt + c
0 = –16t2 + 15t + 2  Substitute 0 for h, 15 for v, and 2 for c.
Step 2: Use the quadratic formula.
x =
–b ± b2 – 4ac 2a
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5. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
(continued)
t =
–15 ± – 4(–16)(2)
2(–16)
Substitute –16 for a, 15 for b, 2 for c, and t for x.
–15 ±
–32
t =
Simplify.
–15 ±
t = –
–32 or
–15 – 18.79
Write two solutions. t
–0.12 or
1.06
Simplify. Use the positive answer because it is the only reasonable answer in this situation.
The ball will land in about 1.06 seconds.
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6. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
Which method(s) would you choose to solve each equation? Justify your reasoning.
a. 5x2 + 8x – 14 = 0
Quadratic formula; the equation cannot be factored easily.
b. 25x2 – 169 = 0
Square roots; there is no x term.
c. x2 – 2x – 3 = 0
Factoring; the equation is easily factorable.
d. x2 – 5x + 3 = 0
Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable.
e. 16x2 – 96x = 0
Quadratic formula; the equation cannot be factored easily and the numbers are large.
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7. Using the Discriminant
ALGEBRA 1 LESSON 9-6
Find the number of solutions of x2 = –3x – 7 using the discriminant.
x2 + 3x + 7 = 0 Write in standard form.
b2 – 4ac = 32 – 4(1)(7) Evaluate the discriminant. Substitute for a, b, and c.
= 9 – 28 Use the order of operations.
= –19 Simplify.
Since –19 < 0, the equation has no solution.
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8. Using the Discriminant
ALGEBRA 1 LESSON 9-6
A football is kicked from a starting height of 3 ft with an initial upward velocity of 40 ft/s. Will the football ever reach a height of 30 ft?
h = –16t2 + vt + c Use the vertical motion formula.
30 = –16t2 + 40t + 3 Substitute 30 for h, 40 for v, and 3 for c.
0 = –16t2 + 40t – 27 Write in standard form.
b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant.
= 1600 – 1728 Use the order of operations.
= –128 Simplify.
The discriminant is negative. The football will never reach a height of 30 ft.
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9. Using the Quadratic Formula
ALGEBRA 1 LESSON 9-6
1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.
2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.
3. Suppose a model rocket is launched from a platform 2 ft above the ground with an initial upward velocity of 100 ft/s. After how many seconds will the rocket hit the ground? Round the solution to the nearest hundredth.
1.5, 4
–2.77, 5.77
6.27 seconds
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10. Using the Discriminant
ALGEBRA 1 LESSON 9-6
Find the number of solutions for each equation.
1. 3x2 – 4x = 7
2. 4x2 = 4x – 1
3. –3x2 + 2x – 12 = 0
4. A ball is thrown from a starting height of 4 ft with an initial upward velocity of 30 ft/s. Is it possible for the ball to reach a height of 18 ft?
two
one
none
yes
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