# Simplify the equations

• Jan 2nd 2010, 02:43 PM
Paymemoney
Simplify the equations
Hi
Need help on the following equations:
Express the following in the form $\displaystyle x+y\sqrt2$ with x and y rational numbers:
1)$\displaystyle (7+5\sqrt2)^{\frac{1}{3}}$

Express the following numbers in the form $\displaystyle x+y\sqrt{n}$ where x and y are rational numbers and n is an integer:
2)$\displaystyle \frac{1}{7+5\sqrt2}$
This is what i have done and i looked in the book's answers but i don't understand how they got the answer to be $\displaystyle -7+5\sqrt5$:
$\displaystyle =\frac{1(7+5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

$\displaystyle =\frac{(7+5\sqrt2)}{(7-5\sqrt2)}$

P.S
• Jan 2nd 2010, 03:01 PM
dkaksl
Not sure about the first one. But I think I got the second one for you.

$\displaystyle \frac{(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

$\displaystyle \frac{(7-5\sqrt2)}{(7)^2-(5\sqrt2)^2}$

$\displaystyle \frac{(7-5\sqrt2)}{49-50}$

$\displaystyle \frac{(7-5\sqrt2)}{-1}$

$\displaystyle -7+5\sqrt2$

Basically, you multiply and divide by the same thing. In the case of making the denominator (lower part of the fraction) simple, you multiply both nominator and denominator with $\displaystyle (7-5\sqrt{2})$

Edit: Wait, you sure the book says $\displaystyle \sqrt5$ for the answer?
• Jan 2nd 2010, 03:02 PM
Gusbob
$\displaystyle (7+5\sqrt2)^{\frac{1}{3}}$

Let $\displaystyle x + y\sqrt2 = ^3\sqrt{7 + 5 \sqrt 2}$

$\displaystyle (x + y\sqrt2)^3 = 7 + 5 \sqrt 2$

Expand the left hand side and collect your terms

__________________________________________________

Express the following numbers in the form $\displaystyle x+y\sqrt{n}$ where x and y are rational numbers and n is an integer:

The answer given by your book is wrong (or you made a typo)

$\displaystyle \frac{1}{7+5\sqrt2}$

$\displaystyle =\frac{1(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$ Expanding the denominator

$\displaystyle =\frac{(7-5\sqrt2)}{(49+ 5\sqrt 2 - 5\sqrt 2 - 50)}$

$\displaystyle = \frac{(7-5\sqrt2)}{-1} = -1(7-5\sqrt2)$

$\displaystyle = -7 + 5\sqrt2$
• Jan 2nd 2010, 03:14 PM
Paymemoney
thanks for the solutions.
• Jan 2nd 2010, 03:16 PM
Stroodle
For the first question:

$\displaystyle x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$

$\displaystyle \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2}$

$\displaystyle \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$

So from this you can now get two equations to solve simultaneously:

$\displaystyle x^3+6xy^2=7$

and $\displaystyle 2y^3+3x^2y=5$

Hence $\displaystyle x=1$ and $\displaystyle y=1$

$\displaystyle \Rightarrow 1+\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$
• Jan 2nd 2010, 03:33 PM
Paymemoney
Quote:

Originally Posted by Gusbob
$\displaystyle (7+5\sqrt2)^{\frac{1}{3}}$

Let $\displaystyle x + y\sqrt2 = ^3\sqrt{7 + 5 \sqrt 2}$

$\displaystyle (x + y\sqrt2)^3 = 7 + 5 \sqrt 2$

Expand the left hand side and collect your terms

i expanded the $\displaystyle (x + y\sqrt2)^3$ then do you substitute $\displaystyle 7 + 5 \sqrt 2$ into
$\displaystyle x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$
• Jan 2nd 2010, 03:38 PM
Stroodle
$\displaystyle x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$

$\displaystyle x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}$

So from here you can get the two equations I posted above.
• Jan 2nd 2010, 03:46 PM
Paymemoney
Quote:

Originally Posted by dkaksl
Not sure about the first one. But I think I got the second one for you.

$\displaystyle \frac{(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

$\displaystyle \frac{(7-5\sqrt2)}{(7)^2-(5\sqrt2)^2}$

$\displaystyle \frac{(7-5\sqrt2)}{49-50}$

$\displaystyle \frac{(7-5\sqrt2)}{-1}$

$\displaystyle -7+5\sqrt2$

Basically, you multiply and divide by the same thing. In the case of making the denominator (lower part of the fraction) simple, you multiply both nominator and denominator with $\displaystyle (7-5\sqrt{2})$

Edit: Wait, you sure the book says $\displaystyle \sqrt5$ for the answer?

yeh i made a typo its meant to be $\displaystyle -7+5\sqrt2$
• Jan 2nd 2010, 03:58 PM
Paymemoney
Quote:

Originally Posted by Stroodle
$\displaystyle x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$

$\displaystyle x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}$

So from here you can get the two equations I posted above.

well the book says the answer is $\displaystyle 1+\sqrt2$
• Jan 2nd 2010, 04:16 PM
Stroodle
Quote:

Originally Posted by Paymemoney
well the book says the answer is $\displaystyle 1+\sqrt2$

That's right. When you solve the two equations you get x=1 and y=1.
• Jan 2nd 2010, 04:35 PM
Paymemoney
ok but i still don't understand how you came up with the two equations from the equation $\displaystyle x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$.

i subsitutuded but it was unsuccessful.
• Jan 2nd 2010, 04:46 PM
Stroodle
When you arrange the equation into the form $\displaystyle x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}$

The coefficient of $\displaystyle \sqrt{2}$ on the left hand side is $\displaystyle 2y^3+3x^2y$ and the coefficient of the $\displaystyle \sqrt{2}$ on the right hand side of the equation is $\displaystyle 5$

So one equation to solve is $\displaystyle 2y^3+3x^2y=5$.

Then the next equation is what is not the coefficient of $\displaystyle \sqrt{2}$ on the left, equal to what is not the coeffient of $\displaystyle \sqrt{2}$ on the right:

$\displaystyle x^3+6xy^2=7$
• Jan 2nd 2010, 05:00 PM
Paymemoney
ok,
firstly i understand how you got the two equations but the part that is confusing me is how did you get this equation:$\displaystyle x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$ to

this equation:$\displaystyle x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$.
• Jan 2nd 2010, 05:28 PM
Stroodle
As posted above:

$\displaystyle x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$

$\displaystyle \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2}$ (from cubing both sides)

$\displaystyle \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$
• Jan 2nd 2010, 08:20 PM
11rdc11
Quote:

Originally Posted by Stroodle
For the first question:

$\displaystyle x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$

$\displaystyle \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2}$

$\displaystyle \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$

So from this you can now get two equations to solve simultaneously:

$\displaystyle x^3+6xy^2=7$

and $\displaystyle 2y^3+3x^2y=5$

Hence $\displaystyle x=1$ and $\displaystyle y=1$

$\displaystyle \Rightarrow 1+\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$

Quick question? How would you go about solving the 2 equations if it would not have been as easy as spotting that x=1 and y=1 works. Would you sub y=mx?