Arithmetic operations problem

Jun 2014
7
0
United Kingdom
Hi all,

I wonder if you could let me know how to resolve c) and d)
maths.png

I could resolve a) and b) by factoring however don't know how to add the values inside the brackets in c) and d) (Thinking)

Many thanks,
Matilde
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
\(\displaystyle \sqrt[3]{7+5\sqrt{2}} = x+y\sqrt{2}\)

Cube both sides:

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

Equating coefficients, we find \(\displaystyle x^3+6xy^2 = 7\) and \(\displaystyle 3x^2y+2y^3 = 5\). Looking quickly, I realize that if \(\displaystyle x=y=1\), the system of equations is solved. Hence, \(\displaystyle \sqrt[3]{7+5\sqrt{2}} = 1+\sqrt{2}\)

For the next one, do the same thing:

\(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = x+y\sqrt{2}\)

Square both sides:

\(\displaystyle \dfrac{11}{2}-3\sqrt{2} = \left(x+y\sqrt{2}\right)^2 = x^2+2xy\sqrt{2}+2y^2\)

Equating coefficients, we have \(\displaystyle x^2+2y^2 = \dfrac{11}{2}\) and \(\displaystyle 2xy = -3\)

Let's solve for \(\displaystyle y\) in the equation on the right: \(\displaystyle y = -\dfrac{3}{2x}\) (we know \(\displaystyle x \neq 0\) because otherwise, \(\displaystyle 2xy=0\neq -3\)).

Plugging that into the first equation:

\(\displaystyle \begin{align*}x^2+2\left(-\dfrac{3}{2x}\right)^2 & = \dfrac{11}{2} \\ x^2 + \dfrac{9}{2x^2} & = \dfrac{11}{2} \\ 2x^4+9 & = 11x^2 \\ 2x^4-11x^2+9 & = 0 \\ (2x^2-9)(x^2-1) & = 0\end{align*}\)

Hence, \(\displaystyle 2x^2-9=0\) or \(\displaystyle x^2-1 = 0\). Since \(\displaystyle x\) must be a rational number, it cannot be the first one, so \(\displaystyle x^2-1=0\). Then \(\displaystyle x = \pm 1\).

Before, we found \(\displaystyle y = -\dfrac{3}{2x}\). So, if \(\displaystyle x=1\), then \(\displaystyle y = -\dfrac{3}{2}\). If \(\displaystyle x=-1\), then \(\displaystyle y = \dfrac{3}{2}\).

So, \(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = \pm \left(1-\dfrac{3}{2}\sqrt{2}\right)\)

Checking on a calculator, you find it is \(\displaystyle \sqrt{\dfrac{11}{2}-3\sqrt{2}} = -1+\dfrac{3}{2}\sqrt{2}\).
 
Last edited:

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Another way to do the first one:

\(\displaystyle \begin{align*}\sqrt[3]{7+5\sqrt{2}} & = \sqrt[3]{1+3\sqrt{2}+6+2\sqrt{2}} \\ & = \sqrt[3]{1+3\sqrt{2}+3\left(\sqrt{2}\right)^2+\left(\sqrt{2}\right)^3} \\ & = \sqrt[3]{\left(1+\sqrt{2}\right)^3} \\ & = 1+\sqrt{2}\end{align*}\)

Here, I just looked for an expression that was a cubic.

I'm sure you can do something similar for the second (find a perfect square).
 
Jun 2014
7
0
United Kingdom
Hi,

By looking at it again I am not entirely sure how you realised that x=y=1 in your first answer?

Also, which computer package allows you to write all the formulas?

The second way to resolve it is quite good. Very simple. Makes me feel a bit stupid (Headbang)

Many thanks
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Hi,

By looking at it again I am not entirely sure how you realised that x=y=1 in your first answer?

Also, which computer package allows you to write all the formulas?

The second way to resolve it is quite good. Very simple. Makes me feel a bit stupid (Headbang)

Many thanks
I wanted to find values for \(\displaystyle x,y\) that would make \(\displaystyle x^3+6xy^2 = 7\). I saw the coefficients were 1 and 6, and 1+6 = 7. So, if x=y=1, that would make that equation true. Then I checked the second equation and saw the coefficients were 2 and 3, and I wanted it to add up to 5, so again, x=y=1 worked. It was just good luck.

For formatting the formulas, just click Reply With Quote on my post above to see the formatting. It is built into the forum. You can download it if you want to type up homework. It is called LaTeX. I personally use TeXclipse (a package for Eclipse). But that is because I like Eclipse for software development, as well.