University Math Challenge?

Question

I have an assignment do monday, and I need these three questions answered.

1)Solve sqrt(1-x) +2x + I x-1 I = 5/4

2)An arithmetic and geometric sequence share the first term of 4 and have the same second term. If the thrid term of the geometric sequence is one more than the third term of the arithmetic sequence, find these sequences.

3)Find all degree values of theta: (cos:theta + isin:theta)^5 = -sqrt(3)/2 + 1/2i

Answer 1

For the first one, solve sqrt(1-x)+2x+ (x-1)=5/4, and also solve sqrt(1-x)+2x - (x-1)=5/4 for x. This is because |x-1|=x-1 or -(x-1).

The way to solve for x is to isolate the square root part, then square both sides and find x. Then you need to plug your possible solutions in and see if they work. From the second equation, you should get x=-5/4.

Answer 2

1)

sqrt(1 - x) + 2x + |x - 1| = 5/4

Your first step would be to deal with that pesky absolute value. Move everything else but the absolute value to the right hand side. This will give you

|x - 1| = (5/4) - sqrt(1 - x) - 2x

This branches off into two equations:

x - 1 = (5/4) - sqrt(1 - x) - 2x, AND
x - 1 = -[(5/4) - sqrt(1 - x) - 2x]

Let's solve these separately.

a) x - 1 = (5/4) - sqrt(1 - x) - 2x

Let's move the -sqrt(1 - x) to the left hand side, and everything else to the right hand side.

sqrt(1 - x) = (5/4) - 2x - x + 1

Combine like terms,

sqrt(1 - x) = (5/4) - 3x + 1

Merge the (5/4) + 1 into (9/4)

sqrt(1 - x) = (9/4) - 3x

And now, make the right hand side into a simple fraction.

sqrt(1 - x) = (9/4) - (12x)/4
sqrt(1 - x) = (9 - 12x)/4

Factoring a 3 on the top,

sqrt(1 - x) = 3(3 - 4x)/4

And now, pulling out the constant (3/4),

sqrt(1 - x) = (3/4) (3 - 4x)

Now, we square both sides.

1 - x = (9/16) (3 - 4x)^2

Expand the squared binomial,

1 - x = (9/16) (9 - 24x + 16x^2)

Multiply both sides by 16,

16 - 16x = 9(9 - 24x + 16x^2)
16 - 16x = 81 - 216x + 144x^2
0 = 144x^2 - 200x + 65

This factors as

x = [200 +/- sqrt(40000 - 37440)] / 288
x = [200 +/- sqrt(2560)] / 288
x = [200 +/- 16sqrt(10)] / 288
x = [25 +/- 2sqrt(10)] / 36

So we get two values for x:
x = {[25 + 2sqrt(10)] / 36, [25 - 2sqrt(10)] / 36}

Both of these values may not work, so they need to be plugged into the equation x - 1 = (5/4) - sqrt(1 - x) - 2x to determine if they are indeed solutions.

b) x - 1 = -[(5/4) - sqrt(1 - x) - 2x]

x - 1 = -5/4 + sqrt(1 - x) + 2x

Isolating the sqrt(1 - x) again,

x - 2x - 1 + 5/4 = sqrt(1 - x)
-3x - 1 + 5/4 = sqrt(1 - x)
-3x + 1/4 = sqrt(1 - x)
(1/4) (-12x + 1) = sqrt(1 - x)

Squaring both sides,

(1/16) (1 - 12x)^2 = 1 - x
(1/16) (1 - 24x + 144x^2) = 1 - x
1 - 24x + 144x^2 = 16 - 16x
144x^2 - 8x - 15 = 0

And using the quadratic formula,

x = [8 +/- sqrt(64 - 4(144)(-15))] / 288
x = [8 +/- sqrt(8704)] / 288
x = [8 +/- 16sqrt(34)] / 288
x = [1 +/- 2sqrt(34)] / 36

So we have two values

x = {[1 + 2sqrt(34)] / 36 , [1 - 2sqrt(34)] / 36}

Again, we can't assume both of these values will work. I'll leave it up to you to plug it back into the equation
x - 1 = -[(5/4) - sqrt(1 - x) - 2x]

2) We have an arithmetic sequence and a geometric sequence. We're given the first term for each of them is 4. Let A denote the arithmetic sequence, and G denote the geometric sequence. So far we have

A = (4, ?, ?)
G = (4, ?, ?)

Arithmetic sequences differ by a difference; that is, if a is the first term, then a + d would be the second term. Geometric sequences differ by a ratio; if a is the first term, then ar would be the next.

A = (4, 4 + d, ?)
G = (4, 4r, ?)

We are given that these sequences have the same second term, so:

4 + d = 4r

Finally, we're given that the third term of the geometric sequence is one more than the third term of the arithmetic sequence. Let's fill in the blanks.


A = (4, 4 + d, 4 + 2d)
G = (4, 4r, 4r^2)

The third term of G is one more than the third term of A, so
4r^2 = (4 + 2d) + 1

So our two equations and two knowns are:

4 + d = 4r
4r^2 = (4 + 2d) + 1

Solving for d in the first equation yields d = 4r - 4.
Plugging this into the second equation means

4r^2 = 4 + 2(4r - 4) + 1
4r^2 = 4 + 8r - 8 + 1
4r^2 = 8r - 3
4r^2 - 8r + 3 = 0

This means

r = [8 +/- sqrt(64 - 4(4)(3))] / 8
r = [8 +/- sqrt(64 - 48)] / 8
r = [8 +/- sqrt(16)]/8
r = [8 +/- 4]/8, so
r = {12/8, 4/8}
r = {3/2, 1/2}

Now that we have two values for r, we will subsequently have two values for d.

If r = 3/2, then
d = 4(3/2) - 4 = 6 - 4 = 2

If r = 1/2, then
d = 4(1/2) - 4 = 2 - 4 = -2

So we have the following pairs of solutions:
{r = 3/2, d = 2}
{r = 1/2, d = -2}

We're going to have two sets of sequences:

A1 = (4, 6, 8)
G1 = (4, 6, 9)

A2 = (4, 2, 0)
G2 = (4, 2, 1)

Answer 3

1)Solve sqrt(1-x) +2x + I x-1 I = 5/4, 1-x≧0 or x≤1

Solution

√(1-x)+2x+1-x = 5/4, 1-x≧0 or x≤1
√(1-x) = 1/4-x, x≤1/4
16(1-x) = 1-8x+16x^2
16x^2+8x-15 = 0
(4x-3)(4x+5) = 0
x = -5/4

Check:
LHS = 3/2-5/2+1+5/4 = 5/4 = RHS
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2)An arithmetic and geometric sequence share the first term of 4 and have the same second term. If the thrid term of the geometric sequence is one more than the third term of the arithmetic sequence, find these sequences.

Solution

4+d = 4r......(1)
4+2d+1 = 4r^2......(2)
(2)-2x(1):
-3 = 4r^2-8r
(2r-1)(2r-3) = 0
r = 1/2, d = -2
or
r = 3/2, d = 2

The two sequences are:
4,2,0,..,4-2(n-1)
4,2,1,...,4(1/2)^(n-1)
or
4,6,8...,4+2(n-1)
4,6,9,...,4(3/2)^(n-1)

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3)Find all degree values of theta: (cos:theta + isin:theta)^5 = -sqrt(3)/2 + 1/2i

Solution

e^(i5ө) = e^(i150+i360n)
ө = (30+72n) degrees , n is any integer.

Answer 4

1) absolute value so for x<0,=-x

sqrt(1-x) +2x -x +1=5/4

(sqrt(1-x))^(2)= (5/4 -x -1)(5/4-x-1)

1-x= 25/16 -5/4x -5/4 -5/4x +x^2+x -5/4 +x +1

1-x= 41/16-10/4x -10/4+x^2

0= x^2+-6/4x-56/16+41/16

0=x^2+-6/4x-15/16

use quadratic formula:

x=

[(6/4)+-sqrt(36/16-4(1)(-15/16))/(2)]

do the seond one yourself too long

Answer 5

sorry... i'v just started to learn my abc's and 123's. i can tell you what 2+2 is... its 6.