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- Thread starter LuluRox
- Start date
LivingMath
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- Jan 29, 2010
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Did you check to see if the answer in the key is correct?
\(\displaystyle \begin{align} 3\sqrt{6-x^2} - \sqrt{14-x^2} &= 0\\3\sqrt{6-{\sqrt{5}}^2} - \sqrt{14-{\sqrt{5}}^2} &= 0\\3\sqrt{6-5} - \sqrt{14-5} &= 0\\3\sqrt{1} - \sqrt{9} &= 0\\3(1) - 3 &= 0\\0&=0\end{align}\)
So, yes, the key is correct. There are often mistakes in answer keys, so always double check their answers if that's not what you're getting.
To solve radical equations, first get rid of all plus and minus signs between radicals (square roots). The goals is to be able to square both sides and eliminate the radicals.
\(\displaystyle \begin{align} 3\sqrt{6-x^2} - \sqrt{14-x^2} &= 0 \\ 3\sqrt{6-x^2} &= \sqrt{14-x^2} \\ (3\sqrt{6-x^2})^2 &= (\sqrt{14-x^2})^2 \\ 9(6-x^2) &= 14-x^2 \end{align}\)
Now just solve for x.
\(\displaystyle \begin{align} 3\sqrt{6-x^2} - \sqrt{14-x^2} &= 0\\3\sqrt{6-{\sqrt{5}}^2} - \sqrt{14-{\sqrt{5}}^2} &= 0\\3\sqrt{6-5} - \sqrt{14-5} &= 0\\3\sqrt{1} - \sqrt{9} &= 0\\3(1) - 3 &= 0\\0&=0\end{align}\)
So, yes, the key is correct. There are often mistakes in answer keys, so always double check their answers if that's not what you're getting.
To solve radical equations, first get rid of all plus and minus signs between radicals (square roots). The goals is to be able to square both sides and eliminate the radicals.
\(\displaystyle \begin{align} 3\sqrt{6-x^2} - \sqrt{14-x^2} &= 0 \\ 3\sqrt{6-x^2} &= \sqrt{14-x^2} \\ (3\sqrt{6-x^2})^2 &= (\sqrt{14-x^2})^2 \\ 9(6-x^2) &= 14-x^2 \end{align}\)
Now just solve for x.
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
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\(\displaystyle Solve \ 3\sqrt{6-x^2}-\sqrt{14-x^2} \ = \ 0\)
\(\displaystyle Hence, \ 3\sqrt{6-x^2} \ = \ \sqrt{14-x^2}\)
\(\displaystyle Squaring \ both \ sides \ gives: \ 9(6-x^2) \ = \ 14-x^2\)
\(\displaystyle 54-9x^2 \ = \ 14-x^2\)
\(\displaystyle 8x^2 \ = \ 40, \ x^2 \ = \ 5, \ \sqrt x^2 \ = \ \sqrt 5, \ |x| \ = \ \sqrt 5, \ x \ = \ \pm\sqrt 5\)
\(\displaystyle Now, \ since \ we \ were \ dealing \ with \ radicals, \ it \ is \ mandatory \ that \ we \ check \ for\)
\(\displaystyle extraneous \ roots, \ however \ there \ aren't \ any, \ so \ the \ solution \ rings \ true.\)
\(\displaystyle Hence, \ 3\sqrt{6-x^2} \ = \ \sqrt{14-x^2}\)
\(\displaystyle Squaring \ both \ sides \ gives: \ 9(6-x^2) \ = \ 14-x^2\)
\(\displaystyle 54-9x^2 \ = \ 14-x^2\)
\(\displaystyle 8x^2 \ = \ 40, \ x^2 \ = \ 5, \ \sqrt x^2 \ = \ \sqrt 5, \ |x| \ = \ \sqrt 5, \ x \ = \ \pm\sqrt 5\)
\(\displaystyle Now, \ since \ we \ were \ dealing \ with \ radicals, \ it \ is \ mandatory \ that \ we \ check \ for\)
\(\displaystyle extraneous \ roots, \ however \ there \ aren't \ any, \ so \ the \ solution \ rings \ true.\)