Simplify the equations

**Paymemoney** · January 2nd 2010, 02:43 PM

Hi
Need help on the following equations:
Express the following in the form $x+y\sqrt2$ with x and y rational numbers:
1) $(7+5\sqrt2)^{\frac{1}{3}}$

Express the following numbers in the form $x+y\sqrt{n}$ where x and y are rational numbers and n is an integer:
2) $\frac{1}{7+5\sqrt2}$
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 $-7+5\sqrt5$ :
$=\frac{1(7+5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

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

P.S

**dkaksl** · January 2nd 2010, 03:01 PM

Not sure about the first one. But I think I got the second one for you.

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

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

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

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

$-7+5\sqrt2$

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 $(7-5\sqrt{2})$

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

**Gusbob** · January 2nd 2010, 03:02 PM

$(7+5\sqrt2)^{\frac{1}{3}}$

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

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

Expand the left hand side and collect your terms

__________________________________________________

Express the following numbers in the form $x+y\sqrt{n}$ 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)

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

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

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

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

$= -7 + 5\sqrt2$

**Paymemoney** · January 2nd 2010, 03:14 PM

thanks for the solutions.

**Stroodle** · January 2nd 2010, 03:16 PM

For the first question:

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

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

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

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

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

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

Hence $x=1$ and $y=1$

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

**Paymemoney** · January 2nd 2010, 03:33 PM

Originally Posted by Gusbob

$(7+5\sqrt2)^{\frac{1}{3}}$

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

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

Expand the left hand side and collect your terms

i expanded the $(x + y\sqrt2)^3$ then do you substitute $7 + 5 \sqrt 2$ into
$x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$

**Stroodle** · January 2nd 2010, 03:38 PM

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

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

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

**Paymemoney** · January 2nd 2010, 03:46 PM

Originally Posted by dkaksl

Not sure about the first one. But I think I got the second one for you.

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

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

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

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

$-7+5\sqrt2$

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 $(7-5\sqrt{2})$

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

yeh i made a typo its meant to be $-7+5\sqrt2$

**Paymemoney** · January 2nd 2010, 03:58 PM

Originally Posted by Stroodle

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

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

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

well the book says the answer is $1+\sqrt2$

**Stroodle** · January 2nd 2010, 04:16 PM

Originally Posted by Paymemoney

well the book says the answer is $1+\sqrt2$

That's right. When you solve the two equations you get x=1 and y=1.

**Paymemoney** · January 2nd 2010, 04:35 PM

ok but i still don't understand how you came up with the two equations from the equation $x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$ .

i subsitutuded but it was unsuccessful.

**Stroodle** · January 2nd 2010, 04:46 PM

When you arrange the equation into the form $x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}$

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

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

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

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

**Paymemoney** · January 2nd 2010, 05:00 PM

ok,
firstly i understand how you got the two equations but the part that is confusing me is how did you get this equation: $x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$ to

this equation: $x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$ .

**Stroodle** · January 2nd 2010, 05:28 PM

As posted above:

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

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

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

**11rdc11** · January 2nd 2010, 08:20 PM

Originally Posted by Stroodle

For the first question:

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

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

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

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

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

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

Hence $x=1$ and $y=1$

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

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?

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