College Algebra problems.

**Dig** · July 17th 2008, 04:44 PM

I have a test tomorrow and I was given the study sheet today. These are summer classes, so things are a bit rushed.

My questions involve solving for real and or imaginary solutions to a quadratic equation using quadratic formula and one involving completing the square to solve an equation.

Problem one states "Find all real solutions."
(3x -4)^5/2 = 32

I start off by distributing to get:
15/2x - 20/2 = 32

I then multiply by two on both sides to receive:
15x - 10 = 64
From here, I am puzzled. I am not sure if I am supposed to complete the square or solve for x or use quadratic formula or what.

If I solve for x, I end up with 74/15.

Problem two states "Find all real and/or imaginary solutions."
3x^2 - 4x + 5 = 0

I use quadratic formula on this problem to get:
4(plus/minus) the square root of (-4)^2 - 4(3)(5)
over
2(3)

So, I am left with:
4(plus/minus) the square root of -44 over 6
then
4(plus/minus) 6.63i over 6

So, my solutions are
10.63i/6 which equals 1.77i
and
-2.63i/6 which equals -0.44i
I am not sure as to whether or not I worked this one out properly.

My third and final problem states "Solve the equation by completing the square."
3x^2 - 12x -9 = 0

First off, I divide the problem by 3 to get:
x^2 - 4x - 3 = 0

Then I add three to both sides and square it off to get:
x^2 - 4x + 4 = 7
and I am left with:
(x - 2)^2 = 7

I square everything off and get:
x - 2 = the square root of 7
which then turns into:
x=2(plus/minus) the square root of seven.

I am confident that I worked this one out correctly, but I am unaware as to whether or not I will need to break this one down further or not to get a real solution set.

Any help is more than appreciated. Thank You.

)

**Jhevon** · July 17th 2008, 04:58 PM

Originally Posted by Dig

I have a test tomorrow and I was given the study sheet today. These are summer classes, so things are a bit rushed.

My questions involve solving for real and or imaginary solutions to a quadratic equation using quadratic formula and one involving completing the square to solve an equation.

Problem one states "Find all real solutions."
(3x -4)^5/2 = 32

I start off by distributing to get:
15/2x - 20/2 = 32

I then multiply by two on both sides to receive:
15x - 10 = 64
From here, I am puzzled. I am not sure if I am supposed to complete the square or solve for x or use quadratic formula or what.

If I solve for x, I end up with 74/15.

huh? did you mean $(3x - 4) \cdot \frac 52 = 32$ ? because what you typed was $(3x - 4)^{5/2} = 32$ or maybe $\frac {(3x - 4)^5}2 = 32$

the moral of the story: type your problems correctly and use parentheses.

Problem two states "Find all real and/or imaginary solutions."
3x^2 - 4x + 5 = 0

I use quadratic formula on this problem to get:
4(plus/minus) the square root of (-4)^2 - 4(3)(5)
over
2(3)

So, I am left with:
4(plus/minus) the square root of -44 over 6
then
4(plus/minus) 6.63i over 6

So, my solutions are
10.63i/6 which equals 1.77i
and
-2.63i/6 which equals -0.44i
I am not sure as to whether or not I worked this one out properly.

yes, the solutions are $x = \frac {4 \pm \sqrt{-44}}6$ which is the same as $\frac 23 \pm \frac { \sqrt{11}}3i$

or if you prefer: $0.67 \pm 1.11i$ (to two decimal places)

My third and final problem states "Solve the equation by completing the square."
3x^2 - 12x -9 = 0

First off, I divide the problem by 3 to get:
x^2 - 4x - 3 = 0

Then I add three to both sides and square it off to get:
x^2 - 4x + 4 = 7
and I am left with:
(x - 2)^2 = 7

I square everything off and get:
x - 2 = the square root of 7
which then turns into:
x=2(plus/minus) the square root of seven.

I am confident that I worked this one out correctly, but I am unaware as to whether or not I will need to break this one down further or not to get a real solution set.

Any help is more than appreciated. Thank You.

)

no, $x = 2 \pm \sqrt{7}$ is fine

good job

**Dig** · July 17th 2008, 05:15 PM

Thank you for your help this far

.

The first problem is
$ (3x - 4)^{5/2} = 32 $

Also, can you elaborate a bit on how you got $ \frac 23 \pm \frac { \sqrt{11}}3i $ from
$ x = \frac {4 \pm \sqrt{-44}}6 $ and $ 0.67 \pm 1.11i $ in the second problem?

I really appreciate your help

**topsquark** · July 17th 2008, 05:29 PM

Originally Posted by Dig

The first problem is
$ (3x - 4)^{5/2} = 32 $

$(3x - 4)^{5/2} = 32$

$\left ( (3x - 4)^{5/2} \right ) ^{2/5} = 32^{2/5}$

$3x - 4 = 32^{2/5} = (2^5)^{2/5} = 2^2 = 4$

$3x - 4 = 4$

I think you can solve this from here. The tough part is to prove that this is the only real solution. Put some effort into how you might find that out. If you are stuck on it, let us know.

-Dan

**topsquark** · July 17th 2008, 05:31 PM

Originally Posted by Dig

Also, can you elaborate a bit on how you got $ \frac 23 \pm \frac { \sqrt{11}}3i $ from
$ x = \frac {4 \pm \sqrt{-44}}6 $ and $ 0.67 \pm 1.11i $ in the second problem?

I really appreciate your help

$-44 = 4 \cdot -11$

So
$\sqrt{-44} = \sqrt{4 \cdot -11} = \sqrt{4} \cdot \sqrt{-11} = 2 \sqrt{-11} = 2i \sqrt{11}$

See if you can get the rest from this.

-Dan

**icemanfan** · July 17th 2008, 05:36 PM

Using $i$ is a way of dealing with solutions that have negative numbers within square roots. To find out the imaginary component of a solution we separate $i$ from the square root. For example, consider the expression $2 + \sqrt{-9}$ . In order to convert this to an acceptable form, we remove $i = \sqrt{-1}$ from the square root. By the laws of multiplication $\sqrt{-9} = \sqrt{-1} \cdot \sqrt{9}$ , which is then equal to $i \cdot 3$ , or $3i$ . Hence our original expression is equal to $2 + 3i$ .

**Dig** · July 17th 2008, 06:31 PM

Alright, thank you all very much for your help thus far.

I understand very thoroughly how you came to this $ \sqrt{-44} = \sqrt{4 \cdot -11} = \sqrt{4} \cdot \sqrt{-11} = 2 \sqrt{-11} = 2i \sqrt{11} $

Now, what I don't understand is how it goes to this $ \frac 23 \pm \frac { \sqrt{11}}3i $ . Why is it being divided by 3?

**icemanfan** · July 17th 2008, 06:37 PM

Originally Posted by Dig

Alright, thank you all very much for your help thus far.

I understand very thoroughly how you came to this $ \sqrt{-44} = \sqrt{4 \cdot -11} = \sqrt{4} \cdot \sqrt{-11} = 2 \sqrt{-11} = 2i \sqrt{11} $

Now, what I don't understand is how it goes to this $ \frac 23 \pm \frac { \sqrt{11}}3i $ . Why is it being divided by 3?

You have the fraction $\frac{4 \pm 2i\sqrt{11}}{6}$ . In reducing that to lowest terms, you remove a factor of 2 from both the numerator and denominator so it becomes $\frac{2 \pm i\sqrt{11}}{3}$ . Then you separate the numerator so that you can make two fractions, one real and one imaginary.

**Dig** · July 17th 2008, 06:50 PM

Originally Posted by icemanfan

You have the fraction $\frac{4 \pm 2i\sqrt{11}}{6}$ . In reducing that to lowest terms, you remove a factor of 2 from both the numerator and denominator so it becomes $\frac{2 \pm i\sqrt{11}}{3}$ . Then you separate the numerator so that you can make two fractions, one real and one imaginary.

Thank you very much my friend. I understand it now. Would the solutions simply be $\frac{2 \pm i\sqrt{11}}{3}$ ?

I have a word problem that I am puzzled by as well. I believe that I am supposed to use quadratic formula on it, but I am not entirely sure. It is as follows.

A ball is thrown downward from a window in a tall building. It's position at time t in seconds is s= 16t^2 + 32t, where s is measured in feet. How long (to the nearest tenth of a second) will it take the ball to fall 262 feet?
To clarify things, the problem reads s is equal to 16t squared plus 32t.

I came up with the answer of 3.17 seconds as my other solution, -5.17, is a negative number.

As stated before, any help is more than appreciated

.

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