Member: TheXSquaredFactor

I broke up each fraction into individual terms, and I knew that each one would, on average, be $\frac{1}{2}$ since $\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2$, so my goal was to make every fraction as close as possible to the value 1/2. If I can do that, then I should be reasonably close to 2.

I filled in the first a/b with 1/2 because that is exactly 1/2, so I have already solved one fraction, but I have used up the 1 and the 2.

Now, a=1 and b=2. That seems pretty good. I then tried this with the other fractions. c/d=3/6 and e/f equals 4/8. I have one fraction left. We now have a few variables defined. They are the following:

a=1

b=2

c=3

d=6

e=4

f=8

The numbers I have left are 5,7, and 9, and there is unfortunately no way of making a fraction that is less than 1/2, so I had to backtrack.

I then realized that I had to make the fractions' average to 1/2, so I decided to take a different approach. I still left a/b=1/2, so a=1 and b=2.

This means that I should have denominators 7,8, and 9. This way, I do not affect the average as much. Let me explain. If I have a fraction with a denominator of 4, for example, then the halfway point is 2. However, the closest possibilities such as 3 and 1 affect the fraction's value by 25%, so I realized that, the larger the denominator, the less the affect is if I increment the numerator by 1.

a=1

b=2

I have now reduced the possibilities to $\frac{1}{2}+\frac{c}{7}+\frac{e}{8}+\frac{g}{9}$. I knew that none of the fractions should have a 6 because that would be too large for the average to work, so I just tested a few possibilities and $\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}$ is the one that worked the best.

TheXSquaredFactorFeb 3, 2018

+2441

It is also possible to use knowledge of difference and sum of squares and cubes to find all the possible solutions.

$x^6=64$

Move 64 to the other side.

$x^6-64=0$

This can be written as a difference of squares.

$\left(x^3\right)^2-8^2=0$

Since this is a difference of squares, factor it as such.

$(x^3+8)(x^3-8)=0$

Notice how both binomials are now difference and sum of cubes. Factor both completely.

$(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0$

Now, set each factor equal to zero and solve. Let's do the easy ones first.

$x+2=0$	$x-2=0$
$x_1=-2$	$x_2=2$

Now, let's tackle the harder ones.

$x^2-2x+4=0\\ a=1\\ b=-2\\ c=4$

$x^2+2x+4=0\\ a=1\\ b=2\\ c=4$

Neither trinomial can be factored, so we will have to rely on other methods. I'll use the quadratic formula in this case.

$x = {-(-2) \pm\sqrt{(-2)^2-4(1)(4)} \over 2(1)}$

$x = {-(2) \pm\sqrt{(2)^2-4(1)(4)} \over 2(1)}$

Simplify both completely.

$x = {2 \pm\sqrt{-12} \over 2}$

$x = {-2 \pm\sqrt{-12} \over 2}$

Now, convert the square root of a negative number to the simplest form in terms in i.

$x=\frac{2\pm 2i\sqrt{3}}{2}$

$x=\frac{-2\pm 2i\sqrt{3}}{2}$

In both cases, the numerator and denominator both have common factors of two.

$x_{3,4}=1\pm i\sqrt{3}$

$x_{5,6}=-1\pm i\sqrt{3}$

Since the original equation is written in the sixth degree, I know that the maximum number of roots equals the degree, or 6. This means I have found every root.

Let's consolidate those roots together.

$x_{1,2}=\pm 2\\ x_{3,4}=1\pm i\sqrt{3}\\ x_{5,6}=-1\pm i\sqrt{3}$

TheXSquaredFactorFeb 3, 2018

+2441

After thinking about this one for quite some time, I think that I have found the magical conjugate: $3^\frac{1}{3}-1$.

First of all, I will convert all the radicals to exponential form.

$\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}$

Now, let's use the magical conjugate.

$\frac{29^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}\frac{3^\frac{1}{3}-1}{3^\frac{1}{3}-1}$	Let's simplify the denominator first just to prove that this conjugate is indeed as magical as I claim.
$\left(1+3^\frac{1}{3}+9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$	Let's do the distributing of every term.
$13^\frac{1}{3}-11+3^\frac{1}{3}3^\frac{1}{3}-13^\frac{1}{3}+9^\frac{1}{3}3^\frac{1}{3}-19^\frac{1}{3}$	This looks pretty menacing. Let's just simplify this monstrosity somewhat.
$3^\frac{1}{3}-1+\left(3^\frac{1}{3}\right)^2-3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-9^\frac{1}{3}$	This still does not look very nice. There is some cancelling that can occur here.
$-1+3^\frac{2}{3}+\left(3^2\right)^\frac{1}{3}*3^\frac{1}{3}-\left(3^2\right)^\frac{1}{3}$	It is wise here to make every exponential term in the same base. This will allow the computation to occur. Let's continue simplifying to see what will happen.
$-1+3^\frac{2}{3}*3^\frac{1}{3}-3^\frac{2}{3}$	Look at that! More exponential terms are cancelling out. In fact, they are all going away!
$-1+3$
$2$

That's a glorious realization to make, huh? The magical conjugate has transformed such a complex denominator into something extraordinarily simple. Let's calculate the numerator now.

$\left(2*9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$	Once again, convert all exponents to common bases and distribute.
$9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}\\ 23^\frac{2}{3}3^\frac{1}{3}-123^\frac{2}{3}$	Notice how the multiplication involves a common base, so you can add the exponents together.
$23-23^\frac{2}{3}$	Let's bring back that denominator of 2!
$\frac{23-23^\frac{2}{3}}{2}$	Of couse, if both terms in the numerator has a factor of 2, like here, then we can simplify further.
$3-3^\frac{2}{3}\\ 3-\sqrt[3]{3^2}\Rightarrow 3-\sqrt[3]{9}$	Since the problem was given in radical form, it is generally considered good form to remain consistent and provide the answer in radical form, too.

TheXSquaredFactorFeb 3, 2018

+2441

It appears as if $\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}$ is the best combination.

By the way, $\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\approx1.99802$

TheXSquaredFactorFeb 2, 2018

+2441

Question 1:

This question definitely suggests a percent increase since the number of books increased. In other words, 15 plus some percent of 15 equals 21 books.

15 + some percent of 15 = 21

Let's let x = some percent because we do not know what that is.

15 + x of 15 = 21

"Of" in mathematics means multiplication.

15+15x=21

Now, solve for x:

$15+15x=21$	Now, isolate x in this equation.
$15x=6$	Divide by 15 from both sides.
$x=\frac{6}{15}=\frac{2}{5}=0.4$	Of course, we want to the percent increase, so we must convert 0.4, currently in decimal format, to a percent. To do that, just multiply by 100 and slap a percent sign behind the number.
$x=0.4\Rightarrow 40\%$	This, of course, is the increase.

Question 2:

The same concept can be used to calculate the percent decrease. I know that this problem involves that since the original amount is greater than the ending amount.

60 cars - some percent of 60 cars = 24 cars

Let's let x equal some percent.

60 cars - x of 60 cars = 24 cars

As aforementioned, "of" is a direct indicator of multiplication in mathematics.

60-60x=24

$60-60x=24$
$-60x=-36$	Divide by -60 on both sides.
$x=\frac{-36}{-60}=\frac{3}{5}=0.6$	Of course, this needs to be a percentage. Let's convert the answer into one.
$x=0.6\Rightarrow 60\%$	Remember that this is a decrease.

Note: I realize that I could have used the formula $\frac{y_2-y_1}{y_1}*100$ to get the percent changed, but I think that the following methods allow you to understand what is occurring; the formula, on the other hand, does not.

TheXSquaredFactorJan 30, 2018

+2441

You are so incredibly close to the right answer. Let me see if I can steer you in the right direction, if I can.

$6.9^a=4.7^b\\ \log\left(6.9^a\right)=\log\left(4.7^b\right)\\ a\log 6.9=b\log4.7\\ \frac{a}{b}=\frac{\log 4.7}{\log 6.9}\approx0.80$

Your answer is wrong because you solved for something other than what the question asked for. The above is the work that you provided. In the above, you solved for $\frac{a}{b}$, but what is the question asking you to solve for? Adjust your solving so that you are solving for the correct expression.

TheXSquaredFactorJan 30, 2018

+2441

The area formula of a trapezoid is given $A=h\left(\frac{b_1+b_2}{2}\right)$

The problem also tells you the following information:

A = unit of area (in a square unit like square feet)

b_1 = one base

b_2= other base

h = perpendicular height

The order in which you substitute the bases are immaterial in this case since addition abides by the commutative and associative properties.

If the questions asks for "how many square feet," this is the unknown. Square feet is indeed a unit of area, so this would be A in the formula above. We don't know what "A" is yet, but we will solve for it. The question directly states the other variables' meaning.

A = unknown area in square feet

b₁= 125ft

b₂= 81ft

h = 75ft

All this information has been nicely synthesized, and we have established the meaning of every single variable in the original formula. Now, we must use the formula to figure out the unknown or the area in square feet:

$A=h\left(\frac{b_1+b_2}{2}\right)$	We know what the variables equal already, so plug them in.
$A=75\text{ft}*\frac{125\text{ft}+81\text{ft}}{2}$	Now it is a matter of simplifying.
$A=75\text{ft}*\frac{206\text{ft}}{2}$
$A=75\text{ft}*103\text{ft}$	It is imperative to remember that multiplying two common units result in that unit squared. It is just like multiplying common variables together.
$A=7725\text{ft}^2$

TheXSquaredFactorJan 30, 2018

+2441

#1) If the price of an item is $130 and the discount is at 60%, this means the price of the sale price should be 60% of the original price, $130.

What is 60% of 130? Well, the "of" in mathematics is an indicator of the multiplication operation, so the problem is effectively 60% * 130.

$60\%*130$	Convert the percentage to a decimal or fraction and simplify the expression.
$60\%=\frac{60}{100}=\frac{3}{5}$	Now, do the multiplication.
$\frac{3}{5}*\frac{130}{1}$	Notice that the 130 and the 5 have a GCF of 5, so we can do some canceling here.
$3*26$	Do the multiplication.
$\$78.00$	This is the sale price of the particular item.

#2) A mark up means that the store is adding to the original price of the item. In this case, $45 with a 35% markup means that the sale price should increase by 35%. In other words, this means:

Sale price = $45 + (35% of 45)

Let's simplify this.

$\$45+35\%\text{ of }45$	"Of" yet again means multiplication.
$45+35*45$	Using some algebra skills, we can make this one multiplication problem.
$45*1.35$
$\$60.75$

TheXSquaredFactorJan 30, 2018

+2441

For factoring, I generally use the "AC-method." The method does not have a formal name, so I will assign it myself! In a quadratic, which is in the form $\textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}$, you first consider factors of $\textcolor{red}{a}\textcolor{green}{c}$ that sum to $\textcolor{blue}{b}$.

In this particular case for $\textcolor{red}{5}x^2+\textcolor{blue}{22}x+\textcolor{green}{8}$, $\textcolor{red}{a}\textcolor{green}{c}=\textcolor{red}{5}*\textcolor{green}{8}=40$ and $\textcolor{blue}{b}=\textcolor{blue}{22}$. We need to find two numbers that multiply to get 40 and add to get 22. Let's list a few factors.

Factors of 40	Factor's Sum
1,40	41
2,20	22
4,10	14
5,8	13

Of the factors of 40 listed, only 2 and 20 are factors that multiply to 40 and add to 22. We can use this information to manipulate the given trinomial.

$5x^2+22x+8\Rightarrow(5x^2+2x)+(20x+8)$	Notice here that I have not actually changed the value of this expression; I have simply changed the way it looks. The next step is to factor via grouping. I've made the groups easier to identify by adding parentheses to the expression. Factor out the GCF from the individual groups.
$x(5x+2)+4(5x+2)$	Notice how a common factor x and 4 both have a common factor of 5x+2. It is possible to combine them together then!
$(x+4)(5x+2)$

Factoring is one of those computations where there are a multitude of ways to arrive at the same answer. This is just the method I generally default to--even if it is not necessarily most efficient.

TheXSquaredFactorJan 27, 2018

+2441

$49x^8-16y^{14}$ can be written as a difference of squares.

$49x^8-16y^{14}\Rightarrow\left(\textcolor{red}{7x^4}\right)^2-\left(\textcolor{blue}{4y^7}\right)^2$	This proves that this expression can be written as a difference of squares. Remember that a difference of squares can be factored using the following rule: $\textcolor{red}{a}^2-\textcolor{blue}{b}^2=(\textcolor{red}{a}+\textcolor{blue}{b})(\textcolor{red}{a}-\textcolor{blue}{b})$
$\hspace{5mm}\textcolor{red}{a}^2\hspace{3mm}-\hspace{5mm}\textcolor{blue}{b}^2\hspace{4mm}=(\hspace{2mm}\textcolor{red}{a}\hspace{2mm}+\hspace{3mm}\textcolor{blue}{b})(\hspace{2mm}\textcolor{red}{a}\hspace{4mm}-\hspace{2mm}\textcolor{blue}{b})\\ \left(\textcolor{red}{7x^4}\right)^2-\left(\textcolor{blue}{4y^7}\right)^2=(\textcolor{red}{7x^4}+\textcolor{blue}{4y^7})(\textcolor{red}{7x^4}-\textcolor{blue}{4y^7})$	Notice how the rule and this particular expression go hand in hand. I introduced colors to ease understanding. At this point, no more can be done.

TheXSquaredFactorJan 27, 2018

Username	TheXSquaredFactor
Score	2441
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Questions	3
Answers	956

\(\frac{29^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}\frac{3^\frac{1}{3}-1}{3^\frac{1}{3}-1}\)	Let's simplify the denominator first just to prove that this conjugate is indeed as magical as I claim.
\(\left(1+3^\frac{1}{3}+9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)\)	Let's do the distributing of every term.
\(13^\frac{1}{3}-11+3^\frac{1}{3}3^\frac{1}{3}-13^\frac{1}{3}+9^\frac{1}{3}3^\frac{1}{3}-19^\frac{1}{3}\)	This looks pretty menacing. Let's just simplify this monstrosity somewhat.
\(3^\frac{1}{3}-1+\left(3^\frac{1}{3}\right)^2-3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-9^\frac{1}{3}\)	This still does not look very nice. There is some cancelling that can occur here.
\(-1+3^\frac{2}{3}+\left(3^2\right)^\frac{1}{3}*3^\frac{1}{3}-\left(3^2\right)^\frac{1}{3}\)	It is wise here to make every exponential term in the same base. This will allow the computation to occur. Let's continue simplifying to see what will happen.
\(-1+3^\frac{2}{3}*3^\frac{1}{3}-3^\frac{2}{3}\)	Look at that! More exponential terms are cancelling out. In fact, they are all going away!
\(-1+3\)
\(2\)

\(60-60x=24\)
\(-60x=-36\)	Divide by -60 on both sides.
\(x=\frac{-36}{-60}=\frac{3}{5}=0.6\)	Of course, this needs to be a percentage. Let's convert the answer into one.
\(x=0.6\Rightarrow 60\%\)	Remember that this is a decrease.

\(A=h\left(\frac{b_1+b_2}{2}\right)\)	We know what the variables equal already, so plug them in.
\(A=75\text{ft}*\frac{125\text{ft}+81\text{ft}}{2}\)	Now it is a matter of simplifying.
\(A=75\text{ft}*\frac{206\text{ft}}{2}\)
\(A=75\text{ft}*103\text{ft}\)	It is imperative to remember that multiplying two common units result in that unit squared. It is just like multiplying common variables together.
\(A=7725\text{ft}^2\)

\(\$45+35\%\text{ of }45\)	"Of" yet again means multiplication.
\(45+35*45\)	Using some algebra skills, we can make this one multiplication problem.
\(45*1.35\)
\(\$60.75\)

\(\left(2*9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)\)	Once again, convert all exponents to common bases and distribute.
\(9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}\\ 23^\frac{2}{3}3^\frac{1}{3}-123^\frac{2}{3}\)	Notice how the multiplication involves a common base, so you can add the exponents together.
\(23-23^\frac{2}{3}\)	Let's bring back that denominator of 2!
\(\frac{23-23^\frac{2}{3}}{2}\)	Of couse, if both terms in the numerator has a factor of 2, like here, then we can simplify further.
\(3-3^\frac{2}{3}\\ 3-\sqrt[3]{3^2}\Rightarrow 3-\sqrt[3]{9}\)	Since the problem was given in radical form, it is generally considered good form to remain consistent and provide the answer in radical form, too.

\(15+15x=21\)	Now, isolate x in this equation.
\(15x=6\)	Divide by 15 from both sides.
\(x=\frac{6}{15}=\frac{2}{5}=0.4\)	Of course, we want to the percent increase, so we must convert 0.4, currently in decimal format, to a percent. To do that, just multiply by 100 and slap a percent sign behind the number.
\(x=0.4\Rightarrow 40\%\)	This, of course, is the increase.

\(60\%*130\)	Convert the percentage to a decimal or fraction and simplify the expression.
\(60\%=\frac{60}{100}=\frac{3}{5}\)	Now, do the multiplication.
\(\frac{3}{5}*\frac{130}{1}\)	Notice that the 130 and the 5 have a GCF of 5, so we can do some canceling here.
\(3*26\)	Do the multiplication.
\(\$78.00\)	This is the sale price of the particular item.

\(5x^2+22x+8\Rightarrow(5x^2+2x)+(20x+8)\)	Notice here that I have not actually changed the value of this expression; I have simply changed the way it looks. The next step is to factor via grouping. I've made the groups easier to identify by adding parentheses to the expression. Factor out the GCF from the individual groups.
\(x(5x+2)+4(5x+2)\)	Notice how a common factor x and 4 both have a common factor of 5x+2. It is possible to combine them together then!
\((x+4)(5x+2)\)

\(49x^8-16y^{14}\Rightarrow\left(\textcolor{red}{7x^4}\right)^2-\left(\textcolor{blue}{4y^7}\right)^2\)	This proves that this expression can be written as a difference of squares. Remember that a difference of squares can be factored using the following rule: \(\textcolor{red}{a}^2-\textcolor{blue}{b}^2=(\textcolor{red}{a}+\textcolor{blue}{b})(\textcolor{red}{a}-\textcolor{blue}{b})\)
\(\hspace{5mm}\textcolor{red}{a}^2\hspace{3mm}-\hspace{5mm}\textcolor{blue}{b}^2\hspace{4mm}=(\hspace{2mm}\textcolor{red}{a}\hspace{2mm}+\hspace{3mm}\textcolor{blue}{b})(\hspace{2mm}\textcolor{red}{a}\hspace{4mm}-\hspace{2mm}\textcolor{blue}{b})\\ \left(\textcolor{red}{7x^4}\right)^2-\left(\textcolor{blue}{4y^7}\right)^2=(\textcolor{red}{7x^4}+\textcolor{blue}{4y^7})(\textcolor{red}{7x^4}-\textcolor{blue}{4y^7})\)	Notice how the rule and this particular expression go hand in hand. I introduced colors to ease understanding. At this point, no more can be done.