Polynomial, Radical Equations & Inequalities. Help needed by Jan. 9,2007!

**Mulya66** · January 3rd 2007, 04:34 PM

Hello all,

I've got a situation and I would greatly appreciate help from anyone who knows Algebra 2 very well (and preferrably someone who could do the provided problems without taking too much of their time). My daughter is a high school student and is struggling with a certain chapter for the past month and neither can my help aid her to do better. She is enrolled in a homeschool program that's located two hours away from our home, so driving up there and over the phone help from the subject teacher is difficult and obviously has had no affect.

I was wondering if anyone could look over the practice test to review with and provide thoroughly worked out answers for myself so I could possibly know what both myself and my daughter has done wrong (I supervise all her homework and how she learns concepts). We also have a deadline, which is Tuesday, January 9, 2007. Anyone who could do all the problems on here that could guide me towards helping improve my daughter and myself would be wonderful!

Thanks in advance!

The test can be found in .pdf (Adobe Reader) format here,
http://www.freewebs.com/smaii/Algebra2%5FTest%5FCh5.pdf (copy and paste it into your browser).

**Soroban** · January 3rd 2007, 04:50 PM

Hello, Mulya66!

The test can be found in .pdf (Adobe Reader) format here,
freewebs.com/smaii/Algebra2%5FTest%5FCh5.pdf (copy and paste it into your browser).

I tried it and got a blank screen with the notation "Restricted Site" at the bottom.

**Mulya66** · January 3rd 2007, 05:02 PM

Originally Posted by Soroban

Hello, Mulya66!

I tried it and got a blank screen with the notation "Restricted Site" at the bottom.

Oh? That's strange, it works just fine in Mozilla Firefox and Internet Explorer. Try downloading it from a file service I tend to use.
http://www2.sendthisfile.com/d.jsp?t...a5s1lo0z64rZlM

Thanks!

**Quick** · January 3rd 2007, 05:10 PM

It works fine for me.

Alright, the properties of exponents are that:
$x^a\times x^b=x^{(a+b)}$

$\left(x^a\right)^b=x^{(a\times b)}$

$x^{-a}=\frac{1}{x^a}$

$\frac{x^a}{x^b}=x^{(a-b)}$

$(ab)^x=a^xb^x$

$\left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}$

Would you like to know why those are the rules?

1. $6x^4\left(2x\right)^3$

Thus: $6x^4\left(2^3x^3\right)$

Then: $6\times x^4 \times 8^{\swarrow^{\text{this is }2^3}}\times x^3$

Use the commutative property of multiplication: $6\times 8\times x^4\times x^3$

Therefore: $48\times x^{(4+3)}$

So the answer is: $\boxed{6x^4\left(2x\right)^3=48x^7}$

**Soroban** · January 3rd 2007, 08:37 PM

Hello, Mulya66!

Thanks, I got it now . . . Here are the last two . . .

19. Solve the system:

. . $\begin{array}{cc}(1)\\(2)\end{array} \begin{array}{cc} 3x - 2y & =\:1 \\ x + 4y & =\:5\end{array}$

I'll use Elimination . . .

$\text{Multiply (1) by 2: }\;\;6x - 4y \:=\:2$

. . . . . . $\text{Add (2): }\quad x + 4y \:=\:5$

. . .and we have: . $7x \:= \:7\quad\Rightarrow\quad\boxed{x \:= \:1}$

Substitute into (2): . $1 + 4y \:=\:5\quad\Rightarrow\quad\boxed{y \:= \:1}$

20. Solve the system:

. . $\begin{array}{ccc}(1)\\(2)\\(3)\end{array} \begin{array}{ccc} 5x + y - z & =\:11 \\ 2x - 2y + z & =\:\text{-}4 \\ x + 2y + z & =\:3\end{array}$

Add (1) and (2): . $7x - \;\: y \:=\:7\;\;(4)$
Add (1) and (3): . $6x + 3y \;=\:3\;\;(5)$

Multiply (4) by 3: . $21x - 3y \:=\:21$
. . . . . . Add (5): . . $6x + 3y \:=\:14$

. . and we have: . $27x \:= \:35\quad\Rightarrow\quad\boxed{ x \:= \:\frac{35}{27}}$

Substitute into (5): . $6\left(\frac{35}{27}\right) + 3y \:=\:14\quad\Rightarrow\quad\boxed{ y \:=\:\frac{56}{27}}$

Substitute into (3): . $\frac{35}{27} + 2\left(\frac{56}{27}\right) + z \:=\:3\quad\Rightarrow\quad\boxed{ z \:=\:-\frac{22}{9}}$

**earboth** · January 3rd 2007, 09:37 PM

Hello,

here is #2:

$\left(\frac{3xy^2}{-x^4 y^3} \right)^{-2}=\left(\frac{-x^4 y^3} {3xy^2} \right)^{2}=\frac{x^8y^6}{9 x^2 y^4}=\frac{x^6y^2}{9}$

EB

**earboth** · January 3rd 2007, 09:42 PM

Hello,

with #3:
1. expand the brackets
2. collect the equal powers(?)

$4x(x-3)-5(3x^2+4x-2)=4x^2-12x-15x^2-20x+10=-11x^2-32x+10$

EB

**earboth** · January 3rd 2007, 09:47 PM

Hello,

Quick has already published a list of properties. With problem #5 you need:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

$\sqrt[3]{27 a^9 b^6}=\left( 3^3 \right)^{\frac{1}{3}} \cdot \left( a^9 \right)^{\frac{1}{3}} \cdot \left( b^6 \right)^{\frac{1}{3}}=3a^3b^2$

#6: $\sqrt{81x^8}=9x^4$

EB

**earboth** · January 3rd 2007, 09:54 PM

Hello,

The radicand is the number under the root sign. If you can factorize the radicand you can calculate partially the root:
$\sqrt{12}=\sqrt{4 \cdot 3}=2 \cdot \sqrt{3}$

#7:
$3 \sqrt{12} - 8 \sqrt{3} + 5 \sqrt{27}=3 \cdot 2 \sqrt{3} - 8 \sqrt{3} + 5 \cdot 3 \sqrt{3}=\sqrt{3}(6-8+15)=13 \sqrt{3}$

EB

**earboth** · January 3rd 2007, 10:02 PM

Hello,

with #8 you can use this formula:

$(a+b)(a-b)=a^2-b^2$

with #8 you get:

$(\sqrt{3}+\sqrt{6})(\sqrt{3} - \sqrt{6})=(\sqrt{3})^2-(\sqrt{6})^2=-3$

EB

**earboth** · January 4th 2007, 12:05 AM

Hello,

#9: In the set of real numbers the square root of a negative number is not defined. To allow the calculation of square roots of negative numbers the imaginary numbers are defined by:

$\sqrt{-1}=i$

$(-4)^{\frac{-3}{2}}=\left( \left( (-4)^{-1} \right)^3 \right)^{\frac{1}{2}}= \frac{1}{\sqrt{(-4)^3}}=\frac{1}{\sqrt{-64}}$

Split -64 into 2 factors: -64 = 64 * (-1):

$\frac{1}{\sqrt{64 \cdot (-1)}}=\frac{1}{\sqrt{64}} \cdot \frac{1}{\sqrt{-1}}=\frac{1}{8i}$

EB

**earboth** · January 4th 2007, 12:08 AM

Hello,

#10: Use the property of #6:

$8^{\frac{4}{3}} \cdot 8^{\frac{1}{3}}=(2^3)^{\frac{4}{3}} \cdot (2^3)^{\frac{1}{3}}=2^4 \cdot 2^1=32$

EB

**earboth** · January 4th 2007, 12:12 AM

Hello,

#11: Use the definition of #9:

1. Expand the brackets
2. Collect real and imaginary numbers:

$(3+4i)(2-7i)=6-21i+8i-28i^2=6-13i+28 \cdot (-1)=34-13i$

EB

**earboth** · January 4th 2007, 12:17 AM

hello,

#12: Use the rule I've mentioned in #8:

$\frac{1-3i}{4+2i} \cdot \frac{4-2i}{4-2i}=\frac{(1-3i)(4-2i)}{16-4i^2}$ $=\frac{4-2i-12i+6i^2}{20}=\frac{-2-14i}{20}=-\frac{1+7i}{10}$

EB

**earboth** · January 4th 2007, 12:28 AM

Hello,

there are a lot of methods to factor a quadratic term. I don't know which one you have to use or which one you are used to use. So I'll give you the result only :
#15:
$6x^2-7x-3=(2x-3)(3x+1)$

#16:
group the terms and factor each term, afterwards you can factor those factored groups:

$2x^3-6x^2+5x-15=2x^2(x-3)+5(x-3)=(x-3)(2x^2+5)$

EB

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