If $\sqrt{27-10\sqrt{2}} = a+b$, where $a$ is a positive integer and $b$ is between $0$ and $1$, what is $\frac{a+b}{a-b}$?
If $\sqrt{27-10\sqrt{2}} = a+b$, where $a$ is a positive integer and $b$ is between $0$ and $1$, what is $\frac{a+b}{a-b}$?
I actually have no idea how to start this question, other than to square the expression and try to find something using a quadratic in a and b. The topic of the unit is "Integer roots of quadratics", if that helps at all. Thank you!
$\endgroup$ 14 Answers
$\begingroup$Denest the nested radical as
$$\sqrt{27-10\sqrt{2}} = \sqrt{27-2\sqrt{50}} = \sqrt{(\sqrt{25}-\sqrt{2})^2} = 5 -\sqrt{2}$$
Thus, $a= 3$, $b= 2-\sqrt2$ and
$$\frac{a+b}{a-b}= \frac{5-\sqrt2}{1+\sqrt2}=(5-\sqrt2) (\sqrt2-1)=6\sqrt2-7$$
$\endgroup$ $\begingroup$$\sqrt {27 - 10\sqrt 2}$
Is a root of $x^4 - 54x^2 + 259 = x^4 - 54x^2 + 23^2$
$x^4 - 54x^2 + 23^2 = (x^2 -10x + 23)(x^2 + 10x + 23)\\ x = \pm 5 \pm \sqrt 2\\ 5 - \sqrt 2 = \sqrt {27 - 10\sqrt 2}$
$a = 3, b = 2 - \sqrt 2$
$\frac{a+b}{a-b} =\frac{5 - \sqrt 2}{1+\sqrt 2}$
$\frac{(5 - \sqrt 2)(-1+\sqrt 2)}{(1+\sqrt 2)(-1+\sqrt 2)}$
$-7+6\sqrt 2$
$\endgroup$ $\begingroup$$$\sqrt{27-10\sqrt{2}}=a+b \approx 3.58578643763$$
You can probably see that this value is between $3$ and $4$ quickly if you take $\sqrt{2} \approx 1.4$. Then $27-10\sqrt{2}\approx 27-14=13$. So we can say that $\sqrt{27-10\sqrt{2}}\approx \sqrt{13}$. So now we know that $a=3$ because you need to pick $a$ as a positive integer such that $b<1$ and therefore we have $b=\sqrt{27-10\sqrt{2}}-3$. We get the final answer:
$$\frac{a+b}{a-b}=\frac{\sqrt{27-10\sqrt{2}}}{6-\sqrt{27-10\sqrt{2}}}=6\sqrt{2}-7$$
$\endgroup$ $\begingroup$You might guess that the square root is $c+d\sqrt 2$ for $c,d$ integer. Square both sides and equate the integer and $\sqrt 2$ parts.
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