G
Guest
Guest
show that the sqraure root of 3+2 times the squareroot of 2 minus the square root of 3-2 times the square root of 2=2
DrToddMatthews
New member
- Joined
- Dec 20, 2005
- Messages
- 11
This is how I approached it.
Hello Allan:
It took me awhile to interpret the expression on the left side of your equals sign; I think you mean this:
sqrt[ 3 + 2*sqrt(2) ] - sqrt[ 3 - 2*sqrt(2) ] = 2
The strategy here is to get rid of the radical expressions; then, after simplifying the result, we show that we end up with an equation that is obviously true. The first step is to separate the radicals (i.e., move one of the radicals to the right side) and then square both sides of the equation.
sqrt[ 3 + 2*sqrt(2) ] = sqrt[ 3 - 2*sqrt(2) ] + 2
Square both sides:
3 + 2*sqrt(2) = 3 - 2*sqrt(2) + 4*sqrt[ 3 - 2*sqrt(2) ] + 4
Again, we have radicals on each side of the equals sign, so we need to simplify and separate them before squaring both sides again. So, let's notice a couple things.
There are constants on both sides of the equals sign. They can be combined on one side. Also, there are two sqrt(2)s on the left side and two negative sqrt(2)s on the right side. We can combine these "like" terms as well; subtract 7 from both sides, and add 2*sqrt(2) to each side:
4*sqrt(2) - 4 = 4*sqrt[ 3 - 2*sqrt(2) ]
Square both sides:
16*2 - 32*sqrt(2) + 16 = 16*[ 3 - 2*sqrt(2) ]
Now simplify. Divide both sides by 16:
2 - 2*sqrt(2) + 1 = 3 - 2*sqrt(2)
Add sqrt(2) to both sides:
2 + 1 = 3
Since each of our steps is a valid algebraic manipulation, the ending equation is equivalent to the beginning equation. The ending equation is obviously true, so the beginning equation must be true as well.
Hello Allan:
It took me awhile to interpret the expression on the left side of your equals sign; I think you mean this:
sqrt[ 3 + 2*sqrt(2) ] - sqrt[ 3 - 2*sqrt(2) ] = 2
The strategy here is to get rid of the radical expressions; then, after simplifying the result, we show that we end up with an equation that is obviously true. The first step is to separate the radicals (i.e., move one of the radicals to the right side) and then square both sides of the equation.
sqrt[ 3 + 2*sqrt(2) ] = sqrt[ 3 - 2*sqrt(2) ] + 2
Square both sides:
3 + 2*sqrt(2) = 3 - 2*sqrt(2) + 4*sqrt[ 3 - 2*sqrt(2) ] + 4
Again, we have radicals on each side of the equals sign, so we need to simplify and separate them before squaring both sides again. So, let's notice a couple things.
There are constants on both sides of the equals sign. They can be combined on one side. Also, there are two sqrt(2)s on the left side and two negative sqrt(2)s on the right side. We can combine these "like" terms as well; subtract 7 from both sides, and add 2*sqrt(2) to each side:
4*sqrt(2) - 4 = 4*sqrt[ 3 - 2*sqrt(2) ]
Square both sides:
16*2 - 32*sqrt(2) + 16 = 16*[ 3 - 2*sqrt(2) ]
Now simplify. Divide both sides by 16:
2 - 2*sqrt(2) + 1 = 3 - 2*sqrt(2)
Add sqrt(2) to both sides:
2 + 1 = 3
Since each of our steps is a valid algebraic manipulation, the ending equation is equivalent to the beginning equation. The ending equation is obviously true, so the beginning equation must be true as well.
Note that
\(\displaystyle \begin{align*}
\mbox{ 3 + 2\sqrt{2} }&=\mbox{ 1 + 2\sqrt{2} + 2 }\\
\\
\\
\\
\\
\\
&=\mbox{ 1 + 2\sqrt{2} + \left(\sqrt{2}\right)^2 }\\
\\
\\
\\
\\
\\
&=\mbox{ (1 + \sqrt{2})^2}\\
\\
\\
\\
\\
\\
\end{align*}\)
Hence
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 + 2\sqrt{2}} }&=\mbox{ \sqrt{(1 + \sqrt{2})^2} }\\
\\
\\
\\
\\
\\
&=\mbox{ 1 + \sqrt{2}}\\
\end{align*}\)
Similarly,
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 - 2\sqrt{2}} }&=\mbox{ \sqrt{(\sqrt{2} - 1)^2} }\\
\\
\\
\\
\\
\\
&=\mbox{ \sqrt{2} - 1}\\
\end{align*}\)
So
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 + 2\sqrt{2}} - \sqrt{3 - 2\sqrt{2}} }&=\mbox{ (1 + \sqrt{2}) - (\sqrt{2} - 1) }\\
\\
\\
\\
\\
\\
&=\mbox{ 2}
\end{align*}\)
\(\displaystyle \begin{align*}
\mbox{ 3 + 2\sqrt{2} }&=\mbox{ 1 + 2\sqrt{2} + 2 }\\
\\
\\
\\
\\
\\
&=\mbox{ 1 + 2\sqrt{2} + \left(\sqrt{2}\right)^2 }\\
\\
\\
\\
\\
\\
&=\mbox{ (1 + \sqrt{2})^2}\\
\\
\\
\\
\\
\\
\end{align*}\)
Hence
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 + 2\sqrt{2}} }&=\mbox{ \sqrt{(1 + \sqrt{2})^2} }\\
\\
\\
\\
\\
\\
&=\mbox{ 1 + \sqrt{2}}\\
\end{align*}\)
Similarly,
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 - 2\sqrt{2}} }&=\mbox{ \sqrt{(\sqrt{2} - 1)^2} }\\
\\
\\
\\
\\
\\
&=\mbox{ \sqrt{2} - 1}\\
\end{align*}\)
So
\(\displaystyle \begin{align*}
\mbox{ \sqrt{3 + 2\sqrt{2}} - \sqrt{3 - 2\sqrt{2}} }&=\mbox{ (1 + \sqrt{2}) - (\sqrt{2} - 1) }\\
\\
\\
\\
\\
\\
&=\mbox{ 2}
\end{align*}\)