Find the root of
rt%7B2%7D%20%20%3D%200" id="TexFormula1" title=" \sqrt{2x { }^{2} } + x + \sqrt{2} = 0" alt=" \sqrt{2x { }^{2} } + x + \sqrt{2} = 0" align="absmiddle" class="latex-formula">
1 answer:
Answer:
The root of x = -2 + √2
Step-by-step explanation:
<u><em> Step(i):-</em></u>
Given ![\sqrt{2x^{2} } + x +\sqrt{2} = 0](https://tex.z-dn.net/?f=%5Csqrt%7B2x%5E%7B2%7D%20%7D%20%2B%20x%20%2B%5Csqrt%7B2%7D%20%3D%200)
⇒ ![\sqrt{2} } ( x^2 ) ^{\frac{1}{2} } + x +\sqrt{2} = 0](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%20%7D%20%28%20x%5E2%20%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%2B%20x%20%2B%5Csqrt%7B2%7D%20%3D%200)
⇒ ![\sqrt{2}x + x +\sqrt{2} = 0](https://tex.z-dn.net/?f=%5Csqrt%7B2%7Dx%20%20%2B%20x%20%2B%5Csqrt%7B2%7D%20%3D%200)
⇒ ![(\sqrt{2} + 1) x = - \sqrt{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B2%7D%20%20%2B%201%29%20x%20%3D%20-%20%5Csqrt%7B2%7D)
<u><em>Step(ii):-</em></u>
Rationalizing
![x = \frac{-\sqrt{2} }{(\sqrt{2} + 1)} X \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%5Csqrt%7B2%7D%20%7D%7B%28%5Csqrt%7B2%7D%20%20%2B%201%29%7D%20X%20%5Cfrac%7B%28%5Csqrt%7B2%7D%20%20-%201%29%7D%7B%28%5Csqrt%7B2%7D%20%20-%201%29%7D)
Apply ( a-b ) ( a+b) = a²-b²
![x = \frac{-\sqrt{2}(\sqrt{2} -1) }{(\sqrt{2})^{2} - 1^{2} )} = \frac{-(2-\sqrt{2} }{2-1}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%5Csqrt%7B2%7D%28%5Csqrt%7B2%7D%20-1%29%20%7D%7B%28%5Csqrt%7B2%7D%29%5E%7B2%7D%20%20%20-%201%5E%7B2%7D%20%29%7D%20%3D%20%5Cfrac%7B-%282-%5Csqrt%7B2%7D%20%7D%7B2-1%7D)
The root of x = -2 + √2
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