Need help in this...
- Thread starter shweta
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My first response to this incomplete question would be - use a calculator.
Please share your work with us.
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You say "solve", but "solving" requires an equation and at least one variable for which one might solve, neither of which is included within your post.
Did you maybe mean "simplify" in some manner? Or "evaluate" maybe?
When you reply, please include a clear listing of your steps and thoughts so far, or else a specification that you are seeking lesson instruction first. Thank you!
Hello, shweta!
Square both sides:
. . \(\displaystyle x^2 \;=\;\left[\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}\right]^2\)
. . \(\displaystyle x^2 \;=\;\left(\sqrt{2+\sqrt{3}}\right)^2 + 2\sqrt{(2+\sqrt{3})(2-\sqrt{3})} + \left(\sqrt{2-\sqrt{3}}\right)^2\)
. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2\sqrt{4-3} \:+\: 2-\sqrt{3}\)
. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2 \:+\: 2 - \sqrt{3}\)
. . \(\displaystyle x^2 \;=\;6\)
Therefore: .\(\displaystyle x \:=\:\sqrt{6}\)
Square both sides:
. . \(\displaystyle x^2 \;=\;\left[\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}\right]^2\)
. . \(\displaystyle x^2 \;=\;\left(\sqrt{2+\sqrt{3}}\right)^2 + 2\sqrt{(2+\sqrt{3})(2-\sqrt{3})} + \left(\sqrt{2-\sqrt{3}}\right)^2\)
. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2\sqrt{4-3} \:+\: 2-\sqrt{3}\)
. . \(\displaystyle x^2 \;=\;2 + \sqrt{3} \:+\: 2 \:+\: 2 - \sqrt{3}\)
. . \(\displaystyle x^2 \;=\;6\)
Therefore: .\(\displaystyle x \:=\:\sqrt{6}\)