If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
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See http://math.stackexchange.com/questions/1380068/mathematics-radical-numbers-problem – lab bhattacharjee Jul 31 '15 at 09:01
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multiply your equation by what you're trying to solve for, and simplify. – Alan Jul 31 '15 at 09:04
4 Answers
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Hint1:
$$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$$
Hint2: $$x^{2}+5-(x^{2}-3)=8$$
Autolatry
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We have
$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2 \sqrt{x^2+5} = 2+ \sqrt{x^2-3}\\ x^2+5 = 4+ x^2-3+4\sqrt{x^2-3}\\ 4 = 4\sqrt{x^2-3}\\ 1 = \sqrt{x^2-3}\\ 1 = x^2 -3\\ x^2 = 4\\ x = \pm 2 $$
So that
$$\sqrt{x^2+5} + \sqrt{x^2-3} = 3 + 1 = 4$$
Kenneth Goodenough
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Notice, we have $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\tag 1$$ let $$\sqrt{x^2+5} + \sqrt{x^2-3} = y\tag 2$$ Now, multiplying both (1) & (2), we get $$(\sqrt{x^2+5} - \sqrt{x^2-3} )(\sqrt{x^2+5} + \sqrt{x^2-3})=2y$$ $$(\sqrt{x^2+5})^2-(\sqrt{x^2-3})^2=2y$$ $$x^2+5-(x^2-3)=2y$$ $$8=2y\implies y=4$$ Hence, $$\bbox[5px, border:2px solid #C0A000]{\sqrt{x^2+5} + \sqrt{x^2-3} = 4}$$
Harish Chandra Rajpoot
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$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\\ \sqrt{x^2+5} = \sqrt{x^2-3} + 2\\ (\sqrt{x^2+5})^2 = (\sqrt{x^2-3})^2 + 2^2\\x^2+5=x^2-3+4-4\sqrt{x^2-3} \\5-1=-4\sqrt{x^2-3}\\\sqrt{x^2-3}=1 \\x^2-3=1\\ \rightarrow x=\pm 2\\\sqrt{x^2+5} + \sqrt{x^2-3}=\sqrt{(\pm2)^2+5} - \sqrt{(\pm2)^2-3}=3+1=4$$
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Khosrotash
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