Hello everyone,
I'm learning math all over again and can't figure out how to solve this problem:
I know the answer to 9+2√18 = √6 + √3 from the back of the book but I don't know how to get there.
Any help is appreciated.
Regards,
Tymon
Hello everyone,
I'm learning math all over again and can't figure out how to solve this problem:
I know the answer to 9+2√18 = √6 + √3 from the back of the book but I don't know how to get there.
Any help is appreciated.
Regards,
Tymon
Hi Tymon!
Effectively you have the equation 9+2√18 = x^{2} and you want to know x.
There is no real straight forward way to do this, but you can use that (a+b)^{2} = (a^{2} + b^{2}) + 2ab.
If a and/or b contains a square root, their squares won't, but 2ab will still contain a square root.
So we make the assumption that 2√18 matches 2ab and 9 matches (a^{2} + b^{2}).
With a bit of trial and error we find that this works if we pick a=√6 and b=√3.
Hello, Tomsten!
$\displaystyle \text{Find the square root of }\,9 + 2\sqrt{18}$
$\displaystyle \text{Answer: }\,{\color{red}\pm(}\sqrt{6} + \sqrt{3}{\color{red})}$
$\displaystyle \text{We have: }\:9 + 2\sqrt{18} \:=\:9 + 2\sqrt{9\!\cdot\!2} \:=\:9 + 2\!\cdot\!\sqrt{9}\!\cdot\!\sqrt{2} \:=\:9+2\!\cdot\!3\!\cdot\!\sqrt{2} \:=\:9+6\sqrt{2}$
$\displaystyle \text{We want }a + b\sqrt{2}\text{ so that: }\,(a+b\sqrt{2})^2 \:=\:9+6\sqrt{2}$
. . . . . . . . . . . . . . . . .$\displaystyle a^2 + 2ab\sqrt{2} + 2b^2 \:=\:9+6\sqrt{2}$
. . . . . . . . . . . . . . $\displaystyle (a^2+2b^2) + (2ab)\sqrt{2} \:=\:9+6\sqrt{2}$
$\displaystyle \text{Equate coefficients: }\:\begin{Bmatrix}a^2+2b^2 &=& 9 & [1] \\ 2ab &=& 6 & [2]\end{Bmatrix}$
$\displaystyle \text{From [2]: }\:2ab \:=\:6 \quad\Rightarrow\quad b \,=\,\tfrac{3}{a}\;\;[3]$
$\displaystyle \text{Substitute into [1]: }\:a^2 + 2\left(\tfrac{3}{a}\right)^2 \:=\:9 \quad\Rightarrow\quad a^2 + \tfrac{18}{a^2} \:=\:9$
. . $\displaystyle a^4 + 18 \:=\:9a^2 \quad\Rightarrow\quad a^4 - 9a^2 + 18 \:=\:0 \quad\Rightarrow\quad (a^2-3)(a^2-6) \:=\:0$
$\displaystyle \text{Hence: }\:\begin{Bmatrix}a^2-3 \:=\:0 & \Rightarrow & a^3 \:=\:3 & \Rightarrow & a \:=\:\pm\sqrt{3} \\ a^2-6\:=\:0 & \Rightarrow & a^2 \:=\:6 &\Rightarrow& a \:=\:\pm\sqrt{6} \end{Bmatrix}$
$\displaystyle \text{We can drop the }\pm\text{ for now.}$
$\displaystyle \text{Substitute into [3]: }\:\begin{Bmatrix}b \,=\,\frac{3}{\sqrt{3}} & \Rightarrow & b \:=\:\sqrt{3} \\ b \:=\:\frac{3}{\sqrt{6}} & \Rightarrow & b \:=\:\frac{\sqrt{6}}{2} \end{Bmatrix}$
$\displaystyle \text{We have: }\:(a,b) \;=\;\begin{Bmatrix}(\sqrt{3},\,\sqrt{3}) \\ \left(\sqrt{6},\,\frac{\sqrt{6}}{2}\right) \end{Bmatrix}$
$\displaystyle \text{Hence: }\:\begin{Bmatrix}a + b\sqrt{2} &=& \sqrt{3} + \sqrt{3}\sqrt{2} &=& \sqrt{3} + \sqrt{6} \\ a+b\sqrt{2} &=& \sqrt{6} + \frac{\sqrt{6}}{2}\sqrt{2} &=& \sqrt{6} + \sqrt{3} \end{Bmatrix}$
$\displaystyle \text{Therefore: }\:\sqrt{9 + 2\sqrt{18}} \;=\;\pm\left(\sqrt{6} + \sqrt{3}\right)$