Expert: Sherman D. - 5/7/2006

Question
1.For each equation, state the values of a, b and c, that would be used in the quadratic formula. Then use the quadrtic formula to solve the equation and estimate the soluton(s) to the nearest tenth.
A.x^2-7x=2
B.2x^2-4x=14
C.3x^2-2x-2=0

2.solve the following equations. if it is possible to factor, do so, if not, use the quadratic formula. if the problem has no solutions, state this.
a.x^2-5x+6=0
b.x^2-x+2=0
c.x^2+4x+2=0
d.2x^2+7x+6=0
e.x^2+x=15-x
f.3x^2+14+8x^2+3x

3. for each equation,state the values of a, b and c, that would be used in the quadratic formula. then, without solving the equation, deyermine the number of solutions by evaluating the determinant(b^2-4ac).
a.x^2+15x+54=0
b.x^2+2x+9=0
c.5x^2-80=0
d.7x^2-8x=0

4.Solve each equation.
a. (r+3)^2-12=24
b. 3(2-h)^2+4=19

5.solve the following equations by any method.
a. x(x-4)=12       b. 3(x+1)^2-21=0
c. 12=4squareroot of x-6     d. (5x+1)^2=3


Answer
Assuming nothing was typed wrong.

1.)

A.)
x^2 - 7x = 2
x^2 - 7x - 2 = 0

a = 1
b = -7
c = -2

x = (-(-7) ± sqrt((-7)^2 - 4(1)(-2)))/(2(1))
x = (7 ± sqrt(49 + 8))/2
x = (7 ± sqrt(57))/2
x = about 7.3 or -.3

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B.)
2x^2 - 4x = 14
2x^2 - 4x - 14 = 0
2(x^2 - 2x - 7) = 0

a = 1
b = -2
c = -7

x = (2 ± sqrt(4 + 28))/2
x = (2 ± sqrt(32))/2
x = (2 ± sqrt(16 * 2))/2
x = (2 ± 4sqrt(2))/2
x = 1 ± 2sqrt(2)
x = about -1.8 and 3.8

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C.)
3x^2 - 2x - 2 = 0

a = 3
b = -2
c = -2

x = (2 ± sqrt(4 + 24))/6
x = (2 ± sqrt(28))/6
x = (2 ± sqrt(4 * 7))/6
x = (2 ± 2sqrt(7))/6
x = (1/3)(1 ± sqrt(7))
x = about -.5 and 1.2

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2.)

a.)
x^2 - 5x + 6 = 0
(x - 3)(x - 2) = 0
x = 3 or 2

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b.)
x^2 - x + 2 = 0
x = (1 ± sqrt(1 - 8))/2
x = (1 ± sqrt(-7))/2
x = (1 ± isqrt(7))/2

The problem has no real solutions.

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c.)
x^2 + 4x + 2 = 0
x = (-4 ± sqrt(16 - 8))/2
x = (-4 ± sqrt(8))/2
x = (-4 ± sqrt(4 * 2))/2
x = (-4 ± 2sqrt(2))/2
x = -2 ± sqrt(2)

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d.)
2x^2 + 7x + 6 = 0
(x + 2)(2x + 3)
x = -2 or (-3/2)

I used www.quickmath.com to factor this

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e.)
x^2 + x = 15 - x
x^2 + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = -5 or 3

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This one has no equal sign, so if you mean
f.)
3x^2 + 14 + 8x^2 = 3x
3x^2 - 3x + 14 + 8x^2 = 0
11x^2 - 3x + 14 = 0

x = (3 ± sqrt(9 - 616))/22
x = (3 ± sqrt(-607))/22
x = (3 ± isqrt(607))/22

This problem has no real solutions. If you meant it like this

3x^2 + 14 = 8x^2 + 3x, then
5x^2 + 3x - 14 = 0
(x + 2)(5x + 7)
x = -2 or (-7/5)

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3.)

a.)
x^2 + 15x + 54 = 0

a = 1
b = 15
c = 54

D = b^2 - 4ac
D = 15^2 - 4(1)(54)
D = 225 - 216
D = 9

There are 2 real solutions

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b.)
x^2 + 2x + 9 = 0

a = 1
b = 2
c = 9

D = 4 - 36
D = -32

There are 2 complex solutions

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c.)
5x^2 - 80 = 0
5(x^2 - 16) = 0

a = 1
b = 0
c = -16

D = -4(-16)(1)
D = 64

There are 2 real solutions

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d.)
7x^2 - 8x = 0

a = 7
b = -8
c = 0

D = (-8)^2 - 4(7)(-8)
D = 64 + 224
D = 288

There are 2 real solutions

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4.)
a.)
(r + 3)^2 - 12 = 24
(r + 3)^2 = 36
r + 3 = ±6
r = -3 ± 6
r = -9 or 3

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b.)
3(2 - h)^2 + 4 = 19
3(2 - h)^2 = 15
(2 - h)^2 = 5
2 - h = ±sqrt(5)
-h = -2 ± sqrt(5)
h = 2 ± sqrt(5)

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5.)
a.)
x(x - 4) = 12    
x^2 - 4x = 12
x^2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
x = 6 or -2

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b.)
3(x + 1)^2 - 21 = 0
3(x + 1)^2 = 21
(x + 1)^2 = 7
x + 1 = ±sqrt(7)
x = -1 ± sqrt(7)

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c.)
12 = 4sqrt(x - 6)    
3 = sqrt(x - 6)
9 = x - 6
x = 15

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d.)
(5x + 1)^2 = 3
5x + 1 = ±sqrt(3)
5x = -1 ± sqrt(3)
x = (1/5)(-1 ± sqrt(3))

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